Let f:(-1,1) rightarrow mathbb{R} be a differ | vedclass

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

Question

(D) Given $g(x) = (f(2f(x)+2))^2$. Applying the chain rule, we differentiate $g(x)$ with respect to $x$: $g^{\prime}(x) = 2(f(2f(x)+2)) \cdot \frac{d}

Vedclass · Accepted Answer

-$4$

Vedclass · Answer

(D) Given $g(x) = (f(2f(x)+2))^2$. Applying the chain rule, we differentiate $g(x)$ with respect to $x$: $g^{\prime}(x) = 2(f(2f(x)+2)) \cdot \frac{d}{dx}(f(2f(x)+2))$ $g^{\prime}(x) = 2(f(2f(x)+2)) \cdot f^{\prime}(2f(x)+2) \cdot \frac{d}{dx}(2f(x)+2)$ $g^{\prime}(x) = 2(f(2f(x)+2)) \cdot f^{\prime}(2f(x)+2) \cdot 2f^{\prime}(x)$ $g^{\prime}(x) = 4 \cdot f(2f(x)+2) \cdot f^{\prime}(2f(x)+2) \cdot f^{\prime}(x)$. Now, substitute $x=0$: $g^{\prime}(0) = 4 \cdot f(2f(0)+2) \cdot f^{\prime}(2f(0)+2) \cdot f^{\prime}(0)$. Given $f(0)=-1$ and $f^{\prime}(0)=1$: $g^{\prime}(0) = 4 \cdot f(2(-1)+2) \cdot f^{\prime}(2(-1)+2) \cdot (1)$ $g^{\prime}(0) = 4 \cdot f(0) \cdot f^{\prime}(0) \cdot 1$ $g^{\prime}(0) = 4 \cdot (-1) \cdot (1) \cdot 1 = -4$.

Let $f:(-1,1) \rightarrow \mathbb{R}$ be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. If $g(x)=(f(2f(x)+2))^2$, then $g^{\prime}(0)=$

Similar Questions

If $8 f(x)+6 f\left(\frac{1}{x}\right)=x+5$ and $y=x^2 f(x)$,then $\frac{d y}{d x}$ at $x=-1$ is

If $f(1)=1$ and $f^{\prime}(1)=3$,then the derivative of $f(f(f(x)))+(f(x))^2$ at $x=1$ is

Let $f_1(x) = e^x, f_2(x) = e^{f_1(x)}, \ldots, f_{n+1}(x) = e^{f_n(x)}$ for all $n \geq 1$. Then for any fixed $n$, $\frac{d}{dx} f_n(x)$ is:

Find the derivative of the function: $(x^{2}+1) \cos x$. (Assume that $a, b, c, d, p, q, r, s$ are fixed non-zero constants and $m, n$ are integers.)

If $\frac{d}{d x}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right] = \left(15 x^p-16 x^q+1\right)(x-1)^{-2}$, then $(p, q)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module