Let f(x) = x - 5. If g(x) = [f(4h(x) + 3)]^2 | vedclass

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

Question

(B) Given $f(x) = x - 5$, so $f'(x) = 1$.$g(x) = [f(4h(x) + 3)]^2$.Using the chain rule, $g'(x) = 2[f(4h(x) + 3)] \cdot f'(4h(x) + 3) \cdot \frac{d}{d

Vedclass · Accepted Answer

$-224$

Vedclass · Answer

(B) Given $f(x) = x - 5$, so $f'(x) = 1$.$g(x) = [f(4h(x) + 3)]^2$.Using the chain rule, $g'(x) = 2[f(4h(x) + 3)] \cdot f'(4h(x) + 3) \cdot \frac{d}{dx}(4h(x) + 3)$.$g'(x) = 2[f(4h(x) + 3)] \cdot 1 \cdot 4h'(x) = 8h'(x) \cdot f(4h(x) + 3)$.At $x = 1$, $g'(1) = 8h'(1) \cdot f(4h(1) + 3)$.Substitute $h(1) = 4$ and $h'(1) = -2$:$g'(1) = 8(-2) \cdot f(4(4) + 3) = -16 \cdot f(19)$.Since $f(x) = x - 5$, $f(19) = 19 - 5 = 14$.$g'(1) = -16 \cdot 14 = -224$.

Let $f(x) = x - 5$. If $g(x) = [f(4h(x) + 3)]^2$ and $h(1) = 4, h'(1) = -2$, then $g'(1) = .....$

Similar Questions

Let $f(x)$ be a polynomial function of the second degree. If $f(1) = f(-1)$ and $a_1, a_2, a_3$ are in $A.P.$,then $f'(a_1), f'(a_2), f'(a_3)$ are in

Find the derivative: $\frac{d}{dx} \left[ \frac{e^{ax}}{\sin(bx + c)} \right]$

If $y = \sin^{-1} \left( x\sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^2} \right)$ and $\frac{dy}{dx} = \frac{1}{2\sqrt{x(1 - x)}} + p$,then $p =$

$f(x)$ is differentiable on $\mathbb{R}$ and $f^{\prime}(m) \neq 0, \,m \in \mathbb{R}$. If $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$,then $m=$

If $f(x)=e^{x} g(x)$,$g(0)=4$,and $g^{\prime}(0)=2$,then $f^{\prime}(0)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module