Differentiation — Flashcards
Telangana Open School (TOSS) · Class 12 · Mathematics
23 flashcards for Differentiation (Telangana Open School (TOSS) Class 12 Mathematics) to test yourself on key terms and facts.
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Find the derivative of \( x^5 \) using the power rule.
Answer
Using the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \) Step 1: Identify \( n = 5 \) Step 2: Apply the formula: \[ \frac{d}{dx}(x^5) = 5x^{5-1} = 5x^4 \] Answer: \( 5x^4 \)…
Differentiate \( \sqrt{x} \) from first principles.
Answer
Let \( f(x) = \sqrt{x} \) By definition: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \] Rationalize the numerator: \[ = \lim_{h \to 0} \frac{\s…
Differentiate \( y = x^3 + 2x^2 - 5x + 7 \)
Answer
Use the sum and difference rule: derivative of sum is sum of derivatives. \[ \frac{d}{dx}(x^3) = 3x^2 \] \[ \frac{d}{dx}(2x^2) = 4x \] \[ \frac{d}{dx}(-5x) = -5 \] \[ \frac{d}{dx}(7) = 0 \quad \text{…
When do you use the product rule? Apply it to \( y = (x^2 + 1)(x^3 - 2) \)
Answer
Use the product rule when differentiating the product of two functions: \( \frac{d}{dx}(uv) = u'v + uv' \) Let \( u = x^2 + 1 \Rightarrow u' = 2x \) Let \( v = x^3 - 2 \Rightarrow v' = 3x^2 \) Appl…
Differentiate \( y = \frac{3x + 1}{x - 2} \) using the quotient rule.
Answer
Quotient Rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \) Let \( u = 3x + 1 \Rightarrow u' = 3 \) Let \( v = x - 2 \Rightarrow v' = 1 \) Apply the formula: \[ \frac{dy}{dx…
What is the derivative of a constant function?
Answer
The derivative of any constant function is zero. For example, if \( f(x) = 5 \), then \( f'(x) = 0 \) This is because there is no change in the function value as \( x \) changes. General Rule: \( \…
Differentiate \( y = (2x + 3)^4 \) using the chain rule.
Answer
Chain Rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) Let \( u = 2x + 3 \Rightarrow \frac{du}{dx} = 2 \) Then \( y = u^4 \Rightarrow \frac{dy}{du} = 4u^3 \) Now apply the chain rule: …
Find the derivative of \( \frac{1}{x^2} \)
Answer
Rewrite the function: \( \frac{1}{x^2} = x^{-2} \) Apply the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \) Here, \( n = -2 \) \[ \frac{d}{dx}(x^{-2}) = -2x^{-3} = -\frac{2}{x^3} \] Answer: \( -\f…
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