If y = f(x^2 + 2) and f'(3) = 5,then frac{dy} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the function $y = f(x^2 + 2)$.Applying the chain rule to differentiate with respect to $x$,we get:$\frac{dy}{dx} = f'(x^2 + 2) \cdot \frac{d

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Given the function $y = f(x^2 + 2)$.Applying the chain rule to differentiate with respect to $x$,we get:$\frac{dy}{dx} = f'(x^2 + 2) \cdot \frac{d}{dx}(x^2 + 2)$$\frac{dy}{dx} = f'(x^2 + 2) \cdot (2x)$Now,evaluate the derivative at $x = 1$:$\left. \frac{dy}{dx} \right|_{x=1} = f'(1^2 + 2) \cdot (2 \cdot 1)$$\left. \frac{dy}{dx} \right|_{x=1} = f'(3) \cdot 2$Given that $f'(3) = 5$,substitute this value into the expression:$\left. \frac{dy}{dx} \right|_{x=1} = 5 \cdot 2 = 10$

If $y = f(x^2 + 2)$ and $f'(3) = 5$,then $\frac{dy}{dx}$ at $x = 1$ is:

Similar Questions

The derivative of $\cosh^{-1} x$ with respect to $\log x$ at $x=5$ is

Differentiate the function with respect to $x$: $\cos (\sin x)$

Let $y = \left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x + \log_{e}(1+x)$ for $x > -1$. Then,at $x = 0$,$\frac{dy}{dx}$ equals:

If $y = t^{4/3} - 3t^{-2/3}$,then $\frac{dy}{dt} = $

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x - 2}{x^2 - 3x + 2}, & x \in R - \{1, 2\} \\ 2, & x = 1 \\ 1, & x = 2 \end{cases}$,then $\lim_{x \rightarrow 2} \frac{f(x) - f(2)}{x - 2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module