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Exercise 10.5 · Q2

Q.If y=f(x2+2)y = f(x^2+2) and f′(3)=5f'(3) = 5, then dydx\dfrac{dy}{dx} at x=1x=1 is

(1) 5
(2) 25
(3) 15
(4) 10
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✓ Free question

Step 1. Differentiate y=f(x2+2)y=f(x^2+2) using the chain rule:

dydx=f′(x2+2)⋅ddx(x2+2)=f′(x2+2)⋅2x\frac{dy}{dx}=f'(x^2+2)\cdot\frac{d}{dx}(x^2+2)=f'(x^2+2)\cdot 2x

Step 2. At x=1x=1: x2+2=1+2=3x^2+2=1+2=3, so

dydx∣x=1=f′(3)⋅2(1)=5⋅2=10\left.\frac{dy}{dx}\right|_{x=1}=f'(3)\cdot 2(1)=5\cdot 2=10

Step 3. Matching against the options, 10 is option (4).

✓Final answer

The correct option is (4) 10

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