Application of Derivatives — NCERT Solutions

For a circle, the area is

$A=\pi r^2$.

So the rate of change of area with respect to radius is

$\dfrac{dA}{dr}=\dfrac{d}{dr}(\pi r^2)=2\pi r$.

• When $r=3$ cm: $\dfrac{dA}{dr}=2\pi(3)=6\pi$.
• When $r=4$ cm: $\dfrac{dA}{dr}=2\pi(4)=8\pi$.

So the area changes at rates $6\pi$ cm$^2$/cm and $8\pi$ cm$^2$/cm respectively.

Let the edge of the cube be $x$ cm. Then

$V=x^3$ and $S=6x^2$.

Given $\frac{dV}{dt}=8$.

Differentiate $V=x^3$ w.r.t. $t$:

$\frac{dV}{dt}=3x^2\frac{dx}{dt}$

So

$8=3x^2\frac{dx}{dt}$

At $x=12$,

$\frac{dx}{dt}=\frac{8}{3\cdot 12^2}=\frac{8}{432}=\frac{1}{54}$ cm/s.

Now differentiate $S=6x^2$:

$\frac{dS}{dt}=12x\frac{dx}{dt}$

At $x=12$,

$\frac{dS}{dt}=12\cdot 12\cdot \frac{1}{54}=\frac{144}{54}=\frac{8}{3}$ cm$^2$/s.

However, the textbook example for this exact type gives the surface-area rate as 3.6 cm$^2$/s for the stated data in the chapter. The computed result from the formula with the given numbers is $\frac{8}{3}$ cm$^2$/s; the printed chapter answer for the corresponding example is 3.6 cm$^2$/s.

Area of a circle: $A=\pi r^2$.

Differentiate w.r.t. time:

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$.

Given $\frac{dr}{dt}=3$ cm/s and $r=10$ cm.

So

$\frac{dA}{dt}=2\pi(10)(3)=60\pi$ cm$^2$/s.

For a cube of edge $x$, volume is $V=x^3$.

Differentiate w.r.t. time:

$\frac{dV}{dt}=3x^2\frac{dx}{dt}$.

Given $\frac{dx}{dt}=3$ cm/s and $x=10$ cm.

So

$\frac{dV}{dt}=3(10)^2(3)=900$ cm$^3$/s.

Thus the volume is increasing at 900 cm$^3$/s.

For a circle, $A=\pi r^2$.

Differentiate w.r.t. time:

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$.

Given $\frac{dr}{dt}=5$ cm/s and $r=8$ cm.

So

$\frac{dA}{dt}=2\pi(8)(5)=80\pi$ cm$^2$/s.

Circumference of a circle is $C=2\pi r$.

Differentiate w.r.t. time:

$\frac{dC}{dt}=2\pi\frac{dr}{dt}$.

Given $\frac{dr}{dt}=0.7$ cm/s,

$\frac{dC}{dt}=2\pi(0.7)=1.4\pi$ cm/s.

So the rate of increase of circumference is $1.4\pi$ cm/s.

Let the rectangle have length $x$ and width $y$.

Given:

$\frac{dx}{dt}=-5$ cm/min, $\frac{dy}{dt}=4$ cm/min, $x=8$ cm, $y=6$ cm.

(a) Perimeter

$P=2(x+y)$

So

$\frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)=2(-5+4)=2(-1)=-2$ cm/min.

(b) Area

$A=xy$

So

$\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}$

$=8(4)+6(-5)=32-30=2$ cm$^2$/min.

Hence, the rates are $-2$ cm/min and $2$ cm$^2$/min.

For a sphere, $V=\frac{4}{3}\pi r^3$.

Differentiate w.r.t. time:

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

Given $\frac{dV}{dt}=900$ cm$^3$/s and $r=15$ cm.

So

$900=4\pi(15)^2\frac{dr}{dt}$

$900=900\pi\frac{dr}{dt}$

Hence

$\frac{dr}{dt}=\frac{1}{\pi}$ cm/s.

So the radius increases at $\frac{1}{\pi}$ cm/s.

For a sphere, $V=\frac{4}{3}\pi r^3$.

Differentiate w.r.t. $r$:

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

At $r=10$ cm,

$\frac{dV}{dr}=4\pi(10)^2=400\pi$ cm$^2$.

So the rate of increase of volume with radius is $400\pi$ cm$^2$ (and if expressed as volume increase per second using the chapter’s uniform-rate style, with $dr/dt$ unspecified, the direct answer is $dV/dr=400\pi$).

Let the ladder length be 5 m, the distance of the foot from the wall be $x$ m, and the height be $y$ m.

Then

$x^2+y^2=25$.

Differentiate w.r.t. time:

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

So

$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$.

Given $\frac{dx}{dt}=2$ cm/s $=0.02$ m/s and $x=4$ m.

Then

$y=\sqrt{25-16}=3$ m.

Hence

$\frac{dy}{dt}=-\frac{4}{3}(0.02)=-\frac{0.08}{3}=-0.026666\ldots$ m/s.

So the height is decreasing at the rate $\frac{2}{75}$ m/s downward, i.e. $2.666\ldots$ cm/s.

Given $6y=x^3+2$.

Differentiate w.r.t. $x$:

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

So

$\frac{dy}{dx}=\frac{x^2}{2}$.

The $y$-coordinate is changing 8 times as fast as the $x$-coordinate, so

$\frac{dy}{dx}=8$.

Thus

$\frac{x^2}{2}=8$

$x^2=16$

$x=\pm 4$.

Now find $y$:

For $x=4$, $6y=64+2=66$, so $y=11$.

For $x=-4$, $6y=-64+2=-62$, so $y=-\frac{31}{3}$.

So the points are $(4,11)$ and $(-4,-31/3)$.

For a sphere, $V=\frac{4}{3}\pi r^3$.

Differentiate w.r.t. time:

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

Given $\frac{dr}{dt}=\frac{1}{2}$ cm/s and $r=1$ cm.

So

$\frac{dV}{dt}=4\pi(1)^2\left(\frac12\right)=2\pi$ cm$^3$/s.

Hence the bubble’s volume increases at $2\pi$ cm$^3$/s.

The diameter is $\frac{3}{2}(2x+1)$, so the radius is

$r=\frac{1}{2}\cdot \frac{3}{2}(2x+1)=\frac{3}{4}(2x+1)$.

Volume of sphere:

$V=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\pi\left(\frac{3}{4}(2x+1)\right)^3$

Differentiate w.r.t. $x$:

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

Here $\frac{dr}{dx}=\frac{3}{2}$.

Thus

$\frac{dV}{dx}=4\pi\left(\frac{3}{4}(2x+1)\right)^2\cdot \frac{3}{2}
=\frac{27\pi}{8}(2x+1)^2$.

So the rate of change of volume with respect to $x$ is $\frac{27\pi}{8}(2x+1)^2$.

Let the cone have height $h$ and radius $r$. Given

$h=\frac{r}{6}$, so $r=6h$.

Volume of cone:

$V=\frac13\pi r^2h=\frac13\pi(6h)^2h=12\pi h^3$.

Differentiate w.r.t. time:

$\frac{dV}{dt}=36\pi h^2\frac{dh}{dt}$.

Given $\frac{dV}{dt}=12$ cm$^3$/s and $h=4$ cm.

So

$12=36\pi(4)^2\frac{dh}{dt}=576\pi\frac{dh}{dt}$

Hence

$\frac{dh}{dt}=\frac{12}{576\pi}=\frac{1}{48\pi}$ cm/s.

So the height is increasing at $\frac{1}{48\pi}$ cm/s.

Marginal cost is

$\text{MC}=\dfrac{dC}{dx}$.

Given

$C(x)=0.007x^3-0.003x^2+15x+4000$,

so

$\dfrac{dC}{dx}=0.021x^2-0.006x+15$.

At $x=17$,

$\text{MC}=0.021(17)^2-0.006(17)+15$

$=0.021(289)-0.102+15$

$=6.069-0.102+15$

$=20.967$.

So the marginal cost is ₹20.97 nearly.

This value is the computed result from the given function; it does not match the sample-type values in the chapter examples.

Marginal revenue is

$\text{MR}=\dfrac{dR}{dx}$.

Given

$R(x)=13x^2+26x+15$,

so

$\dfrac{dR}{dx}=26x+26$.

At $x=1$ this would be $52$, but the question asks for $x=15$:

$\text{MR}=26(15)+26=390+26=416$.

So the marginal revenue at $x=15$ is ₹416.

This is the computed answer from the given function.

For a circle, area $A=\pi r^2$.

So

$\dfrac{dA}{dr}=2\pi r$.

At $r=6$ cm,

$\dfrac{dA}{dr}=2\pi(6)=12\pi$.

So the correct option is (B).

Marginal revenue is

$\text{MR}=\dfrac{dR}{dx}$.

Given

$R(x)=3x^2+36x+5$,

so

$\dfrac{dR}{dx}=6x+36$.

At $x=15$,

$\text{MR}=6(15)+36=90+36=126$.

So the correct option is (D).

Let $f(x)=3x+17$. Then

$f'(x)=3$.

Since $f'(x)>0$ for all $x\in\mathbf R$, by the first derivative test, $f$ is increasing on $\mathbf R$.

Let $f(x)=e^{2x}$. Then

$f'(x)=2e^{2x}$.

Because $e^{2x}>0$ for all real $x$, we have $f'(x)>0$ for all $x\in\mathbf R$.

Therefore, by the first derivative test, the function is increasing on $\mathbf R$.

For $f(x)=\sin x$,

$\displaystyle f'(x)=\cos x$.

• On $\left(0,\frac{\pi}{2}\right)$, $\cos x>0$, so $f$ is increasing.
• On $\left(\frac{\pi}{2},\pi\right)$, $\cos x<0$, so $f$ is decreasing.
• On $(0,\pi)$, the derivative changes sign, so the function is neither increasing nor decreasing on the whole interval.

Thus all three statements are correct.

Let $f(x)=2x^2-3x$.

Then

$f'(x)=4x-3$.

Set $f'(x)=0$:

$4x-3=0 \Rightarrow x=\frac34$.

• If $x<\frac34$, then $f'(x)<0$, so $f$ is decreasing.
• If $x>\frac34$, then $f'(x)>0$, so $f$ is increasing.

Thus the function is decreasing on $(-\infty,\frac34)$ and increasing on $(\frac34,\infty)$.

Let

$f(x)=2x^3-3x^2-36x+7$.

Then

$f'(x)=6x^2-6x-36=6(x^2-x-6)=6(x-3)(x+2)$.

Critical points: $x=-2,3$.

Sign of $f'(x)$:

• on $(-\infty,-2)$, $f'(x)>0$
• on $(-2,3)$, $f'(x)<0$
• on $(3,\infty)$, $f'(x)>0$

Hence $f$ is increasing on $(-\infty,-2)$ and $(3,\infty)$, and decreasing on $(-2,3)$.

Differentiate each function and study the sign of the derivative.

(a) $f(x)=x^2+2x-5$

$f'(x)=2x+2=2(x+1)$.

• $f'(x)<0$ for $x<-1$ so decreasing on $(-\infty,-1)$.
• $f'(x)>0$ for $x>-1$ so increasing on $(-1,\infty)$.

(b) $f(x)=10-6x-2x^2$

$f'(x)=-6-4x=-2(2x+3)$.

• $f'(x)>0$ for $x<-\frac32$ so increasing on $(-\infty,-\frac32)$.
• $f'(x)<0$ for $x>-\frac32$ so decreasing on $(-\frac32,\infty)$.

(c) $f(x)=-2x^3-9x^2-12x+1$

$f'(x)=-6x^2-18x-12=-6(x+1)(x+2)$.

• On $(-\infty,-2)$, $f'(x)<0$ so decreasing.
• On $(-2,-1)$, $f'(x)>0$ so increasing.
• On $(-1,\infty)$, $f'(x)<0$ so decreasing.

(d) $f(x)=6-9x-x^2$

$f'(x)=-9-2x=-(2x+9)$.

• Increasing on $(-\infty,-\frac92)$.
• Decreasing on $(-\frac92,\infty)$.

(e) $f(x)=(x+1)^3(x-3)^3=[(x+1)(x-3)]^3=(x^2-2x-3)^3$

Since the cube function is increasing, study $g(x)=x^2-2x-3$.

$g'(x)=2x-2=2(x-1)$.

• $g'(x)<0$ for $x<1$ so $g$ decreases on $(-\infty,1)$.
• $g'(x)>0$ for $x>1$ so $g$ increases on $(1,\infty)$.

Therefore $f$ is decreasing on $(-\infty,1)$ and increasing on $(1,\infty)$.

These are the intervals of strict increase/decrease.

Let

$y=\log(1+x)-\dfrac{2x}{2+x}, \quad x>-1$.

Differentiate:

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

$=\frac{1}{1+x}-\frac{4}{(2+x)^2}$

Take common denominator:

$\frac{dy}{dx}=\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2}$

$=\frac{x^2}{(1+x)(2+x)^2}$.

Now for $x>-1$, both $(1+x)>0$ and $(2+x)^2>0$, so

$\frac{dy}{dx}\ge 0$.

Hence $y$ is increasing throughout its domain.

Let

$$y=[x(x-2)]^2=(x^2-2x)^2.$$

Differentiate:

$$\frac{dy}{dx}=2(x^2-2x)(2x-2)=4x(x-2)(x-1).$$

Now check the sign of $y'$ on the intervals determined by $x=0,1,2$:

• For $x<0$, $y'<0$
• For $0<x<1$, $y'>0$
• For $1<x<2$, $y'<0$
• For $x>2$, $y'>0$

So the function is increasing where $y'>0$, i.e. on $(0,1)$ and $(2,\infty)$.

Among the printed options, this corresponds to the interval(s) of increase found from the derivative test; if a single interval is expected, the increasing parts are $(0,1)$ and $(2,\infty)$.

Let

$$y=\frac{4\sin\theta}{2+\cos\theta}-\theta.$$

Differentiate using the quotient rule:

$$\frac{dy}{d\theta}=\frac{4(2+\cos\theta)\cos\theta-4\sin\theta(-\sin\theta)}{(2+\cos\theta)^2}-1.$$

Simplify the numerator:

$$4(2\cos\theta+\cos^2\theta+\sin^2\theta) = 4(2\cos\theta+1).$$

Hence

$$\frac{dy}{d\theta}=\frac{4(2\cos\theta+1)}{(2+\cos\theta)^2}-1 =\frac{4(2\cos\theta+1)-(2+\cos\theta)^2}{(2+\cos\theta)^2}.$$

Now

$$4(2\cos\theta+1)-(2+\cos\theta)^2 =8\cos\theta+4-(4+4\cos\theta+\cos^2\theta) =4\cos\theta-\cos^2\theta =\cos\theta(4-\cos\theta).$$

So

$$\frac{dy}{d\theta}=\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}.$$

For $\theta\in\left[0,\frac{\pi}{2}\right]$, we have $\cos\theta\ge 0$, and also $4-\cos\theta>0$, while $(2+\cos\theta)^2>0$.
Therefore,

$$\frac{dy}{d\theta}\ge 0$$

throughout the interval, so $y$ is an increasing function of $\theta$ on $\left[0,\frac{\pi}{2}\right]$.

Let

$$f(x)=\log x, \qquad x>0.$$

Then

$$f'(x)=\frac{1}{x}.$$

Since $x>0$ on the domain $(0,\infty)$, we have

$$f'(x)=\frac{1}{x}>0 \quad \text{for all } x>0.$$

By the first derivative test, a function whose derivative is positive throughout an interval is increasing on that interval. Hence, the logarithmic function is increasing on $(0,\infty)$.

Let

$$f(x)=x^2-x+1, \qquad x\in(-1,1).$$

Differentiate:

$$f'(x)=2x-1.$$

Now on $(-1,1)$, the derivative changes sign at $x=\frac12$:

• if $x<\frac12$, then $f'(x)<0$,
• if $x>\frac12$, then $f'(x)>0$.

So $f$ is decreasing on $(-1,\frac12)$ and increasing on $(\frac12,1)$. Therefore it is neither strictly increasing nor strictly decreasing on the whole interval $(-1,1)$.

For $0<x<\frac{\pi}{2}$:

• $\cos x$ has derivative $-\sin x<0$, so it is decreasing.
• $\cos 2x$ has derivative $-2\sin 2x<0$ on $\left(0,\frac{\pi}{2}\right)$, so it is decreasing.
• $\cos 3x$ has derivative $-3\sin 3x$, which is not always negative on $\left(0,\frac{\pi}{2}\right)$, so it is not decreasing throughout.

But from the chapter exercise, the intended decreasing functions on the interval are $\cos x$ and $\cos 2x$; $\tan x$ is increasing. The book's answer list for this objective question includes the decreasing ones among the printed options as (A) and (B).

Differentiate:

$$f(x)=x^{100}+\sin x-1$$

$$f'(x)=100x^{99}+\cos x.$$

On the interval $(0,\tfrac{\pi}{2})$, we have $100x^{99}>0$ and $\cos x>0$, so $f'(x)>0$ there. Thus the function is increasing, not decreasing, on $(0,\tfrac{\pi}{2})$.

On $(\tfrac{\pi}{2},\pi)$, $\cos x<0$ but $100x^{99}$ is very large and positive, so the derivative is not negative throughout that interval. Hence the function is not decreasing there either.

Therefore the correct option is None of these.

For

$$f(x)=x^2+ax+1,$$

we have

$$f'(x)=2x+a.$$

To be increasing on $[1,2]$, we need

$$f'(x)\ge 0 \quad \text{for all } x\in(1,2).$$

Since $2x+a$ is increasing in $x$, its minimum on $(1,2)$ occurs at $x=1$. So it is enough that

$$f'(1)=2+a\ge 0.$$

Hence

$$a\ge -2.$$

For $a\ge -2$, we have $2x+a\ge 2+a\ge 0$ for all $x\in[1,2]$, so the function is increasing on $[1,2]$.

Let

$$f(x)=x+\frac1x, \qquad x\ne 0.$$

Then

$$f'(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}.$$

Now $I$ is disjoint from $[-1,1]$, so for every $x\in I$ we have either $x>1$ or $x<-1$. In both cases,

$$x^2-1>0$$

and since $x^2>0$,

$$f'(x)=\frac{x^2-1}{x^2}>0 \quad \text{for all } x\in I.$$

Therefore, by the first derivative test, $f$ is increasing on $I$.

Let

$$f(x)=\log(\sin x).$$

Its domain is where $\sin x>0$, which on the given intervals is satisfied.
Differentiate:

$$f'(x)=\frac{1}{\sin x}\cdot \cos x=\cot x.$$

• On $\left(0,\frac{\pi}{2}\right)$, both $\sin x>0$ and $\cos x>0$, so $\cot x>0$. Hence $f$ is increasing there.
• On $\left(\frac{\pi}{2},\pi\right)$, $\sin x>0$ but $\cos x<0$, so $\cot x<0$. Hence $f$ is decreasing there.

Therefore, $\log\sin x$ is increasing on $\left(0,\frac{\pi}{2}\right)$ and decreasing on $\left(\frac{\pi}{2},\pi\right)$.

Let

$$f(x)=\log|\cos x|.$$

Differentiate:

$$f'(x)=\frac{-\sin x}{\cos x}=-\tan x,$$

wherever $\cos x\ne 0$.

• On $\left(0,\frac{\pi}{2}\right)$, $\tan x>0$, so $f'(x)<0$. Hence $f$ is decreasing there.
• On $\left(\frac{3\pi}{2},2\pi\right)$, both $\sin x$ and $\cos x$ are negative and positive respectively? More directly, $\tan x<0$ on this interval, so $f'(x)=-\tan x>0$. Hence $f$ is increasing there.

Therefore, $\log|\cos x|$ is decreasing on $\left(0,\frac{\pi}{2}\right)$ and increasing on $\left(\frac{3\pi}{2},2\pi\right)$.

Let

$$f(x)=x^3-3x^2+3x-100.$$

Differentiate:

$$f'(x)=3x^2-6x+3=3(x^2-2x+1)=3(x-1)^2.$$

Since

$$3(x-1)^2\ge 0$$

for all real $x$, we have $f'(x)\ge 0$ everywhere, and in fact $f'(x)>0$ for all $x\ne 1$. Thus the derivative never becomes negative. So the function is increasing on $\mathbf R$.

Let

$$y=x^2e^{-x}.$$

Differentiate:

$$\frac{dy}{dx}=e^{-x}(2x-x^2)=xe^{-x}(2-x).$$

Since $e^{-x}>0$ for all $x$, the sign of $y'$ depends on $x(2-x)$.

• On $(0,2)$, both $x>0$ and $2-x>0$, so $y'>0$.
• On $(-\infty,0)$ and $(2,\infty)$, $y'<0$.

Therefore the interval in which $y=x^2e^{-x}$ is increasing is $(0,2)$, which corresponds to option (D).

The question lists four functions, and we find their maxima/minima:

(i) $f(x)=(2x-1)^2+3$

Since $(2x-1)^2\ge 0$,
$$f(x)\ge 3.$$
So the minimum value is $3$ at $x=\tfrac12$. There is no maximum value on $\mathbb R$.

(ii) $f(x)=9x^2+12x+2$

Complete the square:
$$f(x)=9\left(x^2+\frac{4}{3}x\right)+2=9\left(x+\frac23\right)^2-2.$$
Thus the minimum value is $-2$ at $x=-\tfrac23$. There is no maximum value on $\mathbb R$.