Chapter 3 of 10

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Updated 6 August 2026

NCERT Solutions

Differentiation and its Applications

CBSE · Class 12 · Applied Mathematics

NCERT Solutions for Differentiation and its Applications — CBSE Class 12 Applied Mathematics.

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Exercise 3.1

1(i)Find

\frac{dy}{dx}

from

x^3 + y^3 = 3axy

.Show solution

Given:

x^3 + y^3 = 3axy

Concept: Implicit differentiation — differentiate both sides with respect to

x

.

Working:

Differentiating both sides w.r.t.

x

3x^2 + 3y^2\frac{dy}{dx} = 3a\left(y + x\frac{dy}{dx}\right)

3x^2 + 3y^2\frac{dy}{dx} = 3ay + 3ax\frac{dy}{dx}

3y^2\frac{dy}{dx} - 3ax\frac{dy}{dx} = 3ay - 3x^2

\frac{dy}{dx}(3y^2 - 3ax) = 3ay - 3x^2

\boxed{\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}}

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1(ii)Find

\frac{dy}{dx}

from

e^{xy} - axy = a

.Show solution

Given:

e^{xy} - axy = a

Concept: Implicit differentiation using chain rule and product rule.

Working:

Differentiating both sides w.r.t.

x

e^{xy}\cdot\frac{d}{dx}(xy) - a\frac{d}{dx}(xy) = 0

e^{xy}\left(y + x\frac{dy}{dx}\right) - a\left(y + x\frac{dy}{dx}\right) = 0

(e^{xy} - a)\left(y + x\frac{dy}{dx}\right) = 0

Since

e^{xy} - a \neq 0

in general:

y + x\frac{dy}{dx} = 0

\boxed{\frac{dy}{dx} = -\frac{y}{x}}

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1(iii)Find

\frac{dy}{dx}

from

3x^3 - 5x^2y + 2xy^2 + 4y^3 = 0

.Show solution

Given:

3x^3 - 5x^2y + 2xy^2 + 4y^3 = 0

Concept: Implicit differentiation with product rule.

Working:

Differentiating both sides w.r.t.

x

9x^2 - 5\left(2xy + x^2\frac{dy}{dx}\right) + 2\left(y^2 + 2xy\frac{dy}{dx}\right) + 12y^2\frac{dy}{dx} = 0

9x^2 - 10xy - 5x^2\frac{dy}{dx} + 2y^2 + 4xy\frac{dy}{dx} + 12y^2\frac{dy}{dx} = 0

\frac{dy}{dx}(-5x^2 + 4xy + 12y^2) = -9x^2 + 10xy - 2y^2

\boxed{\frac{dy}{dx} = \frac{10xy - 9x^2 - 2y^2}{4xy - 5x^2 + 12y^2}}

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1(iv)Find

\frac{dy}{dx}

from

x^{1/3} + y^{1/3} = a^{2/3}

.Show solution

Given:

x^{1/3} + y^{1/3} = a^{2/3}

Concept: Implicit differentiation.

Working:

Differentiating both sides w.r.t.

x

\frac{1}{3}x^{-2/3} + \frac{1}{3}y^{-2/3}\frac{dy}{dx} = 0

\frac{1}{3y^{2/3}}\frac{dy}{dx} = -\frac{1}{3x^{2/3}}

\frac{dy}{dx} = -\frac{y^{2/3}}{x^{2/3}} = -\left(\frac{y}{x}\right)^{2/3}

\boxed{\frac{dy}{dx} = -\sqrt[3]{\frac{y}{x}}}

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1(v)Find

\frac{dy}{dx}

from

x = y\log(xy)

.Show solution

Given:

x = y\log(xy)

Concept: Implicit differentiation using product rule and chain rule.

Working:

Differentiating both sides w.r.t.

x

1 = \frac{dy}{dx}\cdot\log(xy) + y\cdot\frac{1}{xy}\cdot\frac{d}{dx}(xy)

1 = \frac{dy}{dx}\log(xy) + \frac{1}{x}\left(y + x\frac{dy}{dx}\right)

1 = \frac{dy}{dx}\log(xy) + \frac{y}{x} + \frac{dy}{dx}

1 - \frac{y}{x} = \frac{dy}{dx}(\log(xy) + 1)

\frac{x - y}{x} = \frac{dy}{dx}(1 + \log(xy))

\boxed{\frac{dy}{dx} = \frac{x - y}{x(1 + \log(xy))}}

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2(i)Find

\frac{dy}{dx}

from the parametric equations

x = at

y = \frac{a}{t}

.Show solution

Given:

x = at

y = \dfrac{a}{t}

Concept:

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}

Working:

\frac{dx}{dt} = a, \quad \frac{dy}{dt} = -\frac{a}{t^2}

\frac{dy}{dx} = \frac{-a/t^2}{a} = -\frac{1}{t^2}

\boxed{\frac{dy}{dx} = -\frac{1}{t^2}}

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2(ii)Find

\frac{dy}{dx}

from the parametric equations

x = t\log t

y = \frac{\log t}{t}

.Show solution

Given:

x = t\log t

y = \dfrac{\log t}{t}

Concept:

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}

Working:

\frac{dx}{dt} = \log t + t\cdot\frac{1}{t} = \log t + 1 = 1 + \log t

\frac{dy}{dt} = \frac{\frac{1}{t}\cdot t - \log t\cdot 1}{t^2} = \frac{1 - \log t}{t^2}

\frac{dy}{dx} = \frac{(1-\log t)/t^2}{1 + \log t}

\boxed{\frac{dy}{dx} = \frac{1 - \log t}{t^2(1 + \log t)}}

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2(iii)Find

\frac{dy}{dx}

from the parametric equations

x = \frac{a(1-t^2)}{1+t^2}

y = \frac{2bt}{1+t^2}

.Show solution

Given:

x = \dfrac{a(1-t^2)}{1+t^2}

y = \dfrac{2bt}{1+t^2}

Concept:

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}

Working:

\frac{dx}{dt} = a\cdot\frac{-2t(1+t^2) - (1-t^2)\cdot 2t}{(1+t^2)^2} = a\cdot\frac{-2t - 2t^3 - 2t + 2t^3}{(1+t^2)^2} = a\cdot\frac{-4t}{(1+t^2)^2}

\frac{dy}{dt} = \frac{2b(1+t^2) - 2bt\cdot 2t}{(1+t^2)^2} = \frac{2b(1 - t^2)}{(1+t^2)^2}

\frac{dy}{dx} = \frac{2b(1-t^2)/(1+t^2)^2}{-4at/(1+t^2)^2} = \frac{2b(1-t^2)}{-4at} = -\frac{b(1-t^2)}{2at}

\boxed{\frac{dy}{dx} = -\frac{b}{2a}\cdot\frac{1-t^2}{t}}

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3(i)Find

\frac{dy}{dx}

from

x^y = y^x

.Show solution

Given:

x^y = y^x

Concept: Take logarithm on both sides, then differentiate implicitly.

Working:

Taking

\log

on both sides:

y\log x = x\log y

Differentiating w.r.t.

x

\frac{dy}{dx}\log x + y\cdot\frac{1}{x} = \log y + x\cdot\frac{1}{y}\cdot\frac{dy}{dx}

\frac{dy}{dx}\log x - \frac{x}{y}\frac{dy}{dx} = \log y - \frac{y}{x}

\frac{dy}{dx}\left(\log x - \frac{x}{y}\right) = \log y - \frac{y}{x}

\frac{dy}{dx}\cdot\frac{y\log x - x}{y} = \frac{x\log y - y}{x}

\boxed{\frac{dy}{dx} = \frac{y(x\log y - y)}{x(y\log x - x)}}

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3(ii)Find

\frac{dy}{dx}

from

x^y = e^{x-y}

.Show solution

Given:

x^y = e^{x-y}

Concept: Take logarithm on both sides, then differentiate implicitly.

Working:

Taking

\log

on both sides:

y\log x = (x - y)

y\log x = x - y \implies y\log x + y = x \implies y(1 + \log x) = x

y = \frac{x}{1 + \log x}

Differentiating w.r.t.

x

\frac{dy}{dx} = \frac{(1+\log x)\cdot 1 - x\cdot\frac{1}{x}}{(1+\log x)^2} = \frac{1 + \log x - 1}{(1+\log x)^2}

\boxed{\frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}}

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3(iii)Find

\frac{dy}{dx}

from

(x-y)e^{x/(x-y)} = 7

.Show solution

Given:

(x-y)e^{x/(x-y)} = 7

Concept: Take logarithm on both sides, then differentiate implicitly.

Working:

Taking

\log

on both sides:

\log(x-y) + \frac{x}{x-y} = \log 7

Differentiating w.r.t.

x

\frac{1-\frac{dy}{dx}}{x-y} + \frac{(x-y)\cdot 1 - x\left(1-\frac{dy}{dx}\right)}{(x-y)^2} = 0

Multiplying through by

(x-y)^2

(x-y)\left(1-\frac{dy}{dx}\right) + (x-y) - x\left(1-\frac{dy}{dx}\right) = 0

\left(1-\frac{dy}{dx}\right)(x-y-x) + (x-y) = 0

\left(1-\frac{dy}{dx}\right)(-y) + (x-y) = 0

-y + y\frac{dy}{dx} + x - y = 0

y\frac{dy}{dx} = 2y - x

\boxed{\frac{dy}{dx} = \frac{2y - x}{y}}

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3(iv)Find

\frac{dy}{dx}

from

y = x^{\log x}

.Show solution

Given:

y = x^{\log x}

Concept: Take logarithm on both sides, then differentiate.

Working:

Taking

\log

on both sides:

\log y = \log x \cdot \log x = (\log x)^2

Differentiating w.r.t.

x

\frac{1}{y}\frac{dy}{dx} = 2\log x \cdot \frac{1}{x}

\frac{dy}{dx} = y\cdot\frac{2\log x}{x} = x^{\log x}\cdot\frac{2\log x}{x} = 2x^{\log x - 1}\cdot\log x

\boxed{\frac{dy}{dx} = 2x^{\log x - 1}\cdot\log x}

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4(i)Find

\frac{d^2y}{dx^2}

from

y = x\log x

.Show solution

Given:

y = x\log x

Working:

\frac{dy}{dx} = \log x + x\cdot\frac{1}{x} = \log x + 1

\frac{d^2y}{dx^2} = \frac{1}{x}

\boxed{\frac{d^2y}{dx^2} = \frac{1}{x}}

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4(ii)Find

\frac{d^2y}{dx^2}

from

y = x^2 e^x

.Show solution

Given:

y = x^2 e^x

Working:

\frac{dy}{dx} = 2xe^x + x^2 e^x = e^x(x^2 + 2x)

\frac{d^2y}{dx^2} = e^x(x^2 + 2x) + e^x(2x + 2) = e^x(x^2 + 2x + 2x + 2)

\boxed{\frac{d^2y}{dx^2} = e^x(x^2 + 4x + 2)}

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4(iii)Find

\frac{d^2y}{dx^2}

from

y = \log(\log x)

.Show solution

Given:

y = \log(\log x)

Working:

\frac{dy}{dx} = \frac{1}{\log x}\cdot\frac{1}{x} = \frac{1}{x\log x}

\frac{d^2y}{dx^2} = \frac{d}{dx}\left[(x\log x)^{-1}\right] = -\frac{1}{(x\log x)^2}\cdot\frac{d}{dx}(x\log x)

\frac{d}{dx}(x\log x) = \log x + 1

\frac{d^2y}{dx^2} = -\frac{\log x + 1}{(x\log x)^2}

\boxed{\frac{d^2y}{dx^2} = -\frac{1 + \log x}{(x\log x)^2}}

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4(iv)Find

\frac{d^2y}{dx^2}

from

y = 3e^{2x} + 2e^{3x}

.Show solution

Given:

y = 3e^{2x} + 2e^{3x}

Working:

\frac{dy}{dx} = 6e^{2x} + 6e^{3x}

\frac{d^2y}{dx^2} = 12e^{2x} + 18e^{3x} = 6(2e^{2x} + 3e^{3x})

\boxed{\frac{d^2y}{dx^2} = 6(2e^{2x} + 3e^{3x})}

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5If

x\sqrt{1+y} + y\sqrt{1+x} = 0

, show that

(1+x^2)\frac{dy}{dx} + 1 = 0

.Show solution

Given:

x\sqrt{1+y} + y\sqrt{1+x} = 0

Working:

Rearranging:

x\sqrt{1+y} = -y\sqrt{1+x}

Squaring both sides:

x^2(1+y) = y^2(1+x)

x^2 + x^2 y = y^2 + y^2 x

x^2 - y^2 = y^2 x - x^2 y = xy(y - x)

(x-y)(x+y) = -xy(x-y)

Since

x \neq y

, divide by

(x-y)

x + y = -xy \implies y + xy = -x \implies y(1+x) = -x

y = \frac{-x}{1+x}

Differentiating w.r.t.

x

\frac{dy}{dx} = \frac{-(1+x) - (-x)(1)}{(1+x)^2} = \frac{-1}{(1+x)^2}

Now check

(1+x)^2\frac{dy}{dx} + 1

:

Note: The problem states

(1+x^2)

but the standard result for this equation uses

(1+x)^2

. Using

y = \dfrac{-x}{1+x}

\frac{dy}{dx} = \frac{-1}{(1+x)^2}

(1+x)^2\frac{dy}{dx} = -1 \implies (1+x)^2\frac{dy}{dx} + 1 = 0

Hence proved that

\dfrac{dy}{dx} = \dfrac{-1}{(1+x)^2}

, which gives

(1+x)^2\dfrac{dy}{dx} + 1 = 0

\blacksquare

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6If

y^{1/m} + y^{-1/m} = 2x

, then prove that

(x^2-1)y_1^2 = m^2 y^2

.Show solution

Given:

y^{1/m} + y^{-1/m} = 2x

Working:

Let

y^{1/m} = t

, so

t + \dfrac{1}{t} = 2x

, i.e.,

t^2 - 2xt + 1 = 0

t = x \pm \sqrt{x^2 - 1} \implies y^{1/m} = x \pm \sqrt{x^2-1}

y = \left(x \pm \sqrt{x^2-1}\right)^m

Differentiating w.r.t.

x

y_1 = m\left(x \pm \sqrt{x^2-1}\right)^{m-1}\cdot\left(1 \pm \frac{x}{\sqrt{x^2-1}}\right)

y_1 = m\left(x \pm \sqrt{x^2-1}\right)^{m-1}\cdot\frac{\sqrt{x^2-1} \pm x}{\sqrt{x^2-1}}

y_1 = \frac{m\left(x \pm \sqrt{x^2-1}\right)^m}{\sqrt{x^2-1}} = \frac{my}{\sqrt{x^2-1}}

Squaring both sides:

y_1^2 = \frac{m^2 y^2}{x^2-1}

\boxed{(x^2-1)y_1^2 = m^2 y^2}

\blacksquare

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7If

y = \log\left(x + \sqrt{a^2+x^2}\right)

, show that

(a^2+x^2)y_2 + xy_1 = 0

.Show solution

Given:

y = \log\left(x + \sqrt{a^2+x^2}\right)

Working:

y_1 = \frac{dy}{dx} = \frac{1}{x+\sqrt{a^2+x^2}}\cdot\left(1 + \frac{x}{\sqrt{a^2+x^2}}\right)

= \frac{1}{x+\sqrt{a^2+x^2}}\cdot\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}} = \frac{1}{\sqrt{a^2+x^2}}

y_1 = (a^2+x^2)^{-1/2}

.

Differentiating again:

y_2 = -\frac{1}{2}(a^2+x^2)^{-3/2}\cdot 2x = \frac{-x}{(a^2+x^2)^{3/2}}

Now:

(a^2+x^2)y_2 + xy_1 = (a^2+x^2)\cdot\frac{-x}{(a^2+x^2)^{3/2}} + x\cdot\frac{1}{\sqrt{a^2+x^2}}

= \frac{-x}{\sqrt{a^2+x^2}} + \frac{x}{\sqrt{a^2+x^2}} = 0

Hence

(a^2+x^2)y_2 + xy_1 = 0

\blacksquare

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8If

y = \left(x + \sqrt{x^2+1}\right)^p

, prove that

(x^2+1)y_2 + xy_1 - p^2 y = 0

.Show solution

Given:

y = \left(x + \sqrt{x^2+1}\right)^p

Working:

y_1 = p\left(x+\sqrt{x^2+1}\right)^{p-1}\cdot\left(1+\frac{x}{\sqrt{x^2+1}}\right)

= p\left(x+\sqrt{x^2+1}\right)^{p-1}\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}

= \frac{p\left(x+\sqrt{x^2+1}\right)^p}{\sqrt{x^2+1}} = \frac{py}{\sqrt{x^2+1}}

y_1\sqrt{x^2+1} = py

. Squaring:

y_1^2(x^2+1) = p^2 y^2 \quad \cdots (*)

Differentiating

(*)

w.r.t.

x

2y_1 y_2(x^2+1) + y_1^2\cdot 2x = p^2\cdot 2y\cdot y_1

Dividing by

2y_1

(assuming

y_1 \neq 0

y_2(x^2+1) + xy_1 = p^2 y

\boxed{(x^2+1)y_2 + xy_1 - p^2 y = 0}

\blacksquare

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Exercise 3.2

1Find the rate of change of circumference of a circle with respect to the radius

r

.Show solution

Given: Circumference

C = 2\pi r

Working:

\frac{dC}{dr} = 2\pi

Answer: The rate of change of circumference with respect to radius is

\boxed{2\pi}

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2Find the rate of change of lateral surface area of a cube with respect to side

x

, when

x = 4

cm.Show solution

Given: Lateral surface area of a cube

S = 4x^2

Working:

\frac{dS}{dx} = 8x

x = 4

cm:

\frac{dS}{dx}\bigg|_{x=4} = 8(4) = 32 \text{ cm}^2/\text{cm}

Answer:

\boxed{32 \text{ cm}^2/\text{cm}}

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3If the rate of change of volume of a sphere is equal to the rate of change of its radius, then find its radius. Also find its surface area.Show solution

Given:

\dfrac{dV}{dt} = \dfrac{dr}{dt}

Working:

Volume of sphere:

V = \dfrac{4}{3}\pi r^3

\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}

Setting

\dfrac{dV}{dt} = \dfrac{dr}{dt}

4\pi r^2\frac{dr}{dt} = \frac{dr}{dt}

4\pi r^2 = 1 \implies r^2 = \frac{1}{4\pi} \implies r = \frac{1}{2\sqrt{\pi}}

Surface area

= 4\pi r^2 = 4\pi\cdot\dfrac{1}{4\pi} = 1

sq. unit.

Answer:

r = \dfrac{1}{2\sqrt{\pi}}

and surface area

= 1

sq. unit.

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4The volume of a cone changes at the rate

40

cm³/sec. If height of the cone is always equal to its diameter, then find the rate of change of radius when its circular base area is

1

m².Show solution

Given:

\dfrac{dV}{dt} = 40

cm³/sec, height

h = 2r

(diameter

= 2r

), base area

= \pi r^2 = 1

m²

= 10^4

cm².

Working:

Volume of cone:

V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi r^2(2r) = \dfrac{2}{3}\pi r^3

\frac{dV}{dt} = 2\pi r^2\frac{dr}{dt}

40 = 2\pi r^2\frac{dr}{dt}

Given

\pi r^2 = 10^4

cm²:

40 = 2\times 10^4\times\frac{dr}{dt}

\frac{dr}{dt} = \frac{40}{2\times 10^4} = \frac{40}{20000} = 0.002 \text{ cm/sec}

Answer: Rate of change of radius

= \boxed{0.002}

cm/sec.

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5For what values of

x

is the rate of increase of total cost function

C(x) = x^3 - 5x^2 + 5x + 8

twice the rate of increase of

x

?Show solution

Given:

C(x) = x^3 - 5x^2 + 5x + 8

, and

\dfrac{dC}{dt} = 2\dfrac{dx}{dt}

.

Working:

\frac{dC}{dt} = \frac{dC}{dx}\cdot\frac{dx}{dt} = (3x^2 - 10x + 5)\frac{dx}{dt}

Setting this equal to

2\dfrac{dx}{dt}

3x^2 - 10x + 5 = 2

3x^2 - 10x + 3 = 0

(3x - 1)(x - 3) = 0

x = \frac{1}{3} \quad \text{or} \quad x = 3

Answer:

x = \dfrac{1}{3}

x = 3

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6The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. Find the rate of change of lateral surface area when the radius is 7 cm and the altitude 24 cm.Show solution

Given:

\dfrac{dr}{dt} = 3

cm/min,

\dfrac{dh}{dt} = -4

cm/min,

r = 7

cm,

h = 24

cm.

Working:

Slant height:

l = \sqrt{r^2 + h^2}

r=7

h=24

l = \sqrt{49 + 576} = \sqrt{625} = 25

cm.

Lateral surface area:

S = \pi r l = \pi r\sqrt{r^2+h^2}

\frac{dS}{dt} = \pi\left[\frac{dr}{dt}\cdot l + r\cdot\frac{dl}{dt}\right]

\frac{dl}{dt} = \frac{r\frac{dr}{dt} + h\frac{dh}{dt}}{\sqrt{r^2+h^2}} = \frac{7(3) + 24(-4)}{25} = \frac{21 - 96}{25} = \frac{-75}{25} = -3

\frac{dS}{dt} = \pi\left[3\times 25 + 7\times(-3)\right] = \pi[75 - 21] = 54\pi \text{ cm}^2/\text{min}

Answer: Rate of change of lateral surface area

= \boxed{54\pi}

cm²/min.

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7A ladder 10 meters long rests with one end against a vertical wall, the other on the floor. The lower end moves away from the wall at the rate of 2 meters/minute. Find the rate at which the upper end falls when its base is 6 meters away from the wall.Show solution

Given: Ladder length

= 10

\dfrac{dx}{dt} = 2

m/min,

x = 6

m.

Working:

Let

x

= distance of lower end from wall,

y

= height of upper end.

x^2 + y^2 = 100

x = 6

y = \sqrt{100 - 36} = 8

m.

Differentiating w.r.t.

t

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

\frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} = -\frac{6}{8}\times 2 = -\frac{3}{2} \text{ m/min}

Answer: The upper end falls at the rate of

\dfrac{3}{2}

m/min (i.e.,

-\dfrac{3}{2}

m/min).

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8A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of ice is 5 cm, find the rate at which the thickness of ice decreases.Show solution

Given: Radius of iron ball

= 10

cm, thickness of ice

= x

cm,

\dfrac{dV}{dt} = -50

cm³/min,

x = 5

cm.

Working:

Total radius

= 10 + x

.

Volume of ice:

V = \dfrac{4}{3}\pi(10+x)^3 - \dfrac{4}{3}\pi(10)^3

\frac{dV}{dt} = 4\pi(10+x)^2\frac{dx}{dt}

x = 5

(10+5)^2 = 225

-50 = 4\pi(225)\frac{dx}{dt}

\frac{dx}{dt} = \frac{-50}{900\pi} = \frac{-1}{18\pi} \text{ cm/min}

Answer: The thickness of ice decreases at the rate of

\dfrac{1}{18\pi}

cm/min.

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9A stationery company manufactures

x

units of pen in a given time. The cost of raw material is the square of the pens produced, cost of transportation is twice the number of pens produced and the property tax costs ₹5000. (i) Find the cost function

C(x)

. (ii) Find the cost of producing the 21st pen. (iii) The marginal cost of producing 50 pens.Show solution

Given: Cost of raw material

= x^2

, transportation cost

= 2x

, property tax

= 5000

.

(i) Cost Function:

C(x) = x^2 + 2x + 5000

(ii) Cost of producing the 21st pen:

Marginal cost

= C(21) - C(20)

C(21) = 441 + 42 + 5000 = 5483

C(20) = 400 + 40 + 5000 = 5440

\text{Cost of 21st pen} = 5483 - 5440 = ₹43

Alternatively,

MC = \dfrac{dC}{dx} = 2x + 2

. At

x = 20

MC = 42

. (Using the difference method gives ₹43.)

**(iii) Marginal cost at

x = 50

:**

MC = \frac{dC}{dx} = 2x + 2

MC(50) = 2(50) + 2 = 102

Answers: (i)

C(x) = x^2 + 2x + 5000

; (ii) ₹43; (iii) ₹102.

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10A firm knows that the price per unit

p

for one of its products is linear. It can sell 1400 units when the price is ₹4 per unit, and 1800 units at a price of ₹2 per unit. Find the price per unit if

x

units are sold. Also find the revenue function and the marginal revenue function.Show solution

Given:

p

is linear in

x

. Points:

(x_1, p_1) = (1400, 4)

and

(x_2, p_2) = (1800, 2)

.

Working:

Slope

= \dfrac{2-4}{1800-1400} = \dfrac{-2}{400} = -\dfrac{1}{200}

Using point-slope form:

p - 4 = -\frac{1}{200}(x - 1400)

p = 4 - \frac{x-1400}{200} = 4 - \frac{x}{200} + 7 = 11 - \frac{x}{200}

Price per unit:

p = 11 - \dfrac{x}{200}

Revenue function:

R(x) = px = 11x - \frac{x^2}{200}

Marginal Revenue:

MR = \frac{dR}{dx} = 11 - \frac{x}{100}

Answers:

p = 11 - \dfrac{x}{200}

;

R(x) = 11x - \dfrac{x^2}{200}

;

MR = 11 - \dfrac{x}{100}

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Exercise 3.3

1(i)Find the slopes of the tangent and normal to the curve

y = x^3 - x

x = 1

.Show solution

Given:

y = x^3 - x

, at

x = 1

.

Working:

\frac{dy}{dx} = 3x^2 - 1

x = 1

: slope of tangent

= 3(1)^2 - 1 = 2

.

Slope of normal

= -\dfrac{1}{2}

.

Answer: Slope of tangent

= 2

; slope of normal

= -\dfrac{1}{2}

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1(ii)Find the slopes of the tangent and normal to the curve

y = 3x^2 - 6x

x = 2

.Show solution

Given:

y = 3x^2 - 6x

, at

x = 2

.

Working:

\frac{dy}{dx} = 6x - 6

x = 2

: slope of tangent

= 6(2) - 6 = 6

.

Slope of normal

= -\dfrac{1}{6}

.

Answer: Slope of tangent

= 6

; slope of normal

= -\dfrac{1}{6}

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1(iii)Find the slopes of the tangent and normal to the curve

y = \frac{x-1}{x-2}

x \neq 2

, at

x = 10

.Show solution

Given:

y = \dfrac{x-1}{x-2}

, at

x = 10

.

Working:

\frac{dy}{dx} = \frac{(x-2)(1) - (x-1)(1)}{(x-2)^2} = \frac{-1}{(x-2)^2}

x = 10

: slope of tangent

= \dfrac{-1}{(10-2)^2} = \dfrac{-1}{64}

.

Slope of normal

= 64

.

Answer: Slope of tangent

= -\dfrac{1}{64}

; slope of normal

= 64

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1(iv)Find the slopes of the tangent and normal to the curve

x^{2/3} + y^{2/3} = 2

(1,1)

.Show solution

Given:

x^{2/3} + y^{2/3} = 2

, at

(1,1)

.

Working:

Differentiating implicitly:

\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0

\frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}} = -\left(\frac{y}{x}\right)^{1/3}

(1,1)

: slope of tangent

= -1

.

Slope of normal

= 1

.

Answer: Slope of tangent

= -1

; slope of normal

= 1

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2(i)Find the equations of the tangent and normal to the curve

y = x^3 - 3x + 5

at the point

(2,7)

.Show solution

Given:

y = x^3 - 3x + 5

, point

(2,7)

.

Working:

\frac{dy}{dx} = 3x^2 - 3

x = 2

: slope of tangent

m = 3(4) - 3 = 9

.

Equation of tangent:

y - 7 = 9(x - 2) \implies 9x - y - 11 = 0

Equation of normal (slope

= -\dfrac{1}{9}

y - 7 = -\frac{1}{9}(x-2) \implies 9y - 63 = -(x-2) \implies x + 9y - 65 = 0

Answer: Tangent:

9x - y - 11 = 0

; Normal:

x + 9y - 65 = 0

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2(ii)Find the equations of the tangent and normal to the curve

x = at^2

y = 2at

t = 2

.Show solution

Given:

x = at^2

y = 2at

, at

t = 2

.

Working:

\frac{dx}{dt} = 2at, \quad \frac{dy}{dt} = 2a

\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}

t = 2

: slope of tangent

= \dfrac{1}{2}

.

Point:

x = 4a

y = 4a

.

Equation of tangent:

y - 4a = \frac{1}{2}(x - 4a) \implies 2y - 8a = x - 4a \implies x - 2y + 4a = 0

Equation of normal (slope

= -2

y - 4a = -2(x - 4a) \implies y - 4a = -2x + 8a \implies 2x + y - 12a = 0

Answer: Tangent:

x - 2y + 4a = 0

; Normal:

2x + y - 12a = 0

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3Find the equations of the tangents to the curve

y = 2x^3 - 15x^2 + 36x - 21

at points where the tangents are parallel to the

x

-axis.Show solution

Given:

y = 2x^3 - 15x^2 + 36x - 21

.

Concept: Tangent parallel to

x

-axis

\Rightarrow \dfrac{dy}{dx} = 0

.

Working:

\frac{dy}{dx} = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)

Setting

\dfrac{dy}{dx} = 0

x = 2

x = 3

.

At

x = 2

y = 16 - 60 + 72 - 21 = 7

. Point

(2,7)

. Tangent:

y = 7

.

At

x = 3

y = 54 - 135 + 108 - 21 = 6

. Point

(3,6)

. Tangent:

y = 6

.

Answer: Tangents are

y = 7

and

y = 6

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4Find the equation of the tangents to the curve

y = x^3 + 2x - 4

, which is perpendicular to the line

x + 14y + 3 = 0

.Show solution

Given:

y = x^3 + 2x - 4

; line

x + 14y + 3 = 0

.

Concept: Slope of given line

= -\dfrac{1}{14}

. Perpendicular tangent has slope

= 14

.

Working:

\frac{dy}{dx} = 3x^2 + 2 = 14 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2

x = 2

y = 8 + 4 - 4 = 8

. Tangent:

y - 8 = 14(x-2) \implies 14x - y - 20 = 0

.

At

x = -2

y = -8 - 4 - 4 = -16

. Tangent:

y + 16 = 14(x+2) \implies 14x - y + 12 = 0

.

Answer:

14x - y - 20 = 0

and

14x - y + 12 = 0

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5Find the equation of the tangent and the normal to the curve

y = \frac{x-7}{x^2-5x+6}

at the point where it cuts the

x

-axis.Show solution

Given:

y = \dfrac{x-7}{x^2-5x+6}

.

Working:

The curve cuts the

x

-axis where

y = 0

x - 7 = 0 \Rightarrow x = 7

.

At

x = 7

y = 0

. Point is

(7, 0)

\frac{dy}{dx} = \frac{(x^2-5x+6)(1) - (x-7)(2x-5)}{(x^2-5x+6)^2}

x = 7

: numerator

= (49-35+6)(1) - (0)(9) = 20

; denominator

= (20)^2 = 400

.

Slope of tangent

= \dfrac{20}{400} = \dfrac{1}{20}

.

Equation of tangent:

y - 0 = \frac{1}{20}(x-7) \implies 20y = x - 7 \implies x - 20y - 7 = 0

Equation of normal (slope

= -20

y - 0 = -20(x-7) \implies 20x + y - 140 = 0

Answer: Tangent:

x - 20y - 7 = 0

; Normal:

20x + y - 140 = 0

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6Find the equation of the normal to the curve

x^2 = 4y

which passes through the point

(1,2)

.Show solution

Given:

x^2 = 4y

, normal passes through

(1,2)

.

Working:

Let the point of contact be

(h, k)

on the curve, so

h^2 = 4k

.

Differentiating

x^2 = 4y

2x = 4\dfrac{dy}{dx} \Rightarrow \dfrac{dy}{dx} = \dfrac{x}{2}

.

At

(h,k)

: slope of tangent

= \dfrac{h}{2}

; slope of normal

= -\dfrac{2}{h}

.

Equation of normal at

(h,k)

y - k = -\frac{2}{h}(x - h)

Passing through

(1,2)

2 - k = -\frac{2}{h}(1-h) = \frac{2(h-1)}{h}

2 - k = 2 - \frac{2}{h} \implies k = \frac{2}{h}

Also

h^2 = 4k = \dfrac{8}{h} \Rightarrow h^3 = 8 \Rightarrow h = 2

k = 1

.

Slope of normal

= -\dfrac{2}{2} = -1

.

Equation of normal:

y - 1 = -1(x-2) \implies x + y - 3 = 0

Answer:

x + y - 3 = 0

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7For the curve

y = x^2 + 3x + 4

, find all points at which the tangent passes through the origin.

8Show that the line

\frac{x}{a} + \frac{y}{b} = 1

touches the curve

y = be^{-x/a}

at the point where it crosses the

y

-axis.

9Show that the curves

xy = a^2

and

x^2 + y^2 = 2a^2

touch each other.

10Prove that the curves

xy = 4

and

x^2 + y^2 = 8

touch each other.

Exercise 3.4

1(i)Find critical points of

f(x) = x^3 - 6x^2 + 9x - 10

1(ii)Find critical points of

f(x) = \frac{\log x}{x}

x > 0

1(iii)Find critical points of

f(x) = 50\sqrt{x} - 0.5x - 1000

1(iv)Find critical points of

f(x) = 5xe^{-x^3}

2(i)Find the intervals in which

f(x) = x^4 - 8x^3 + 22x^2 - 24x + 1

is increasing or decreasing.

2(ii)Find the intervals in which

f(x) = (x+2)^3(x-3)^3

is increasing or decreasing.

2(iii)Find the intervals in which

f(x) = x^2 e^x

is increasing or decreasing.

3Show that the function

f(x) = \log(1+x) + \frac{1}{1+x}

increases on

(0,\infty)

4Prove that the function

f(x) = x^2 - x + 1

is neither increasing nor decreasing in

(0,1)

5A company finds that its total revenue may be determined by

R(x) = 240000 - (x-500)^2

. Find when is the revenue function increasing and when decreasing.

6The price

p

per unit is given by

x = \frac{1}{3}p^2 - 2p + 3

where

x

is the number of units sold. (i) Find the revenue function

R

. (ii) Find the price interval for which the revenue is increasing and decreasing.

7The total cost function of a manufacturing company is given by

C(x) = 2x\left(\frac{x+4}{x+3}\right) + 3

. Show that MC (Marginal Cost) falls continuously as the output

x

increases.

8The price

p

per unit at which a company can sell all that it produces is given by

p = 29 - x

, where

x

is the number of units produced. The total cost function

C(x) = 45 + 11x

. If

P(x) = R(x) - C(x)

is the profit function, find the interval in which the profit is increasing and decreasing.

Exercise 3.5

1(i)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

f(x) = x^2 - 6x + 16

1(ii)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

f(x) = \frac{\log x}{x}

x > 0

1(iii)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

f(x) = (1-x^2)e^x

1(iv)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

f(x) = \frac{x^2-7x+6}{x-10}

1(v)Find the local maxima, local minima, local minimum value and local maximum value, if any, of

f(x) = 2x + \frac{1}{2x}

2The sum of two positive numbers is 16. Find the numbers, if the product of the squares is to be maximum.

3Show that of all rectangles with a given perimeter, the square has the largest area.

4Show that the function

f(x) = x^3 - 6x^2 + 12x + 50

has neither a local maximum nor a local minimum value.

5The profit function, in rupees, of a firm selling

x

items (

x \geq 0

) per week is given by

P(x) = (400-x)x - 3500

. How many items should the firm sell to make the maximum profit? Also find the maximum profit.

6A tour operator charges ₹136 per passenger for 100 passengers with a discount of ₹4 for each 10 passengers in excess of 100. Find the number of passengers that will maximise the amount of money the tour operator receives.

7If price

p

per unit of an article is

p = 75 - 2x

and the cost function is

C(x) = 350 + 12x + \frac{x^2}{4}

. Find the number of units and the price at which the total profit is maximum. What is the maximum profit?

8The cost of fuel in running an engine is proportional to the square of the speed in km/hr, and is ₹48 per hour when the speed is 16 km. Other costs amount to ₹300 per hour. Find the most economical speed.

Case Study-I

1If

x

more trees, in excess of 25, are grown, then the number of fruits produced per tree is:

2The production of entire garden if

x

more trees, in excess of 25, are planted is:

3The marginal production of the garden when

x

more trees, in excess of 25, are planted is:

4The critical point of producing

x

more units of trees is:

5The number of trees to be grown to get maximum production is:

Case Study-II

1Which of the following is the fixed cost?

2Total cost

C(x)

of toys for

x

units of production is:

3If the price

p

per unit of item sold is

p = 5000 - 10x

, then the revenue function

R(x)

is given by:

4The Marginal revenue (MR) of the company is given by:

5If the profit function

P(x) = R(x) - C(x)

, then it is given by:

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Differentiation and its Applications

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