Suppose that h(x) = f(x) cdot g(x) and F(x) = | vedclass

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

Question

(B) Given $h(x) = f(x)g(x)$. Using the product rule,$h'(x) = f'(x)g(x) + g'(x)f(x)$.At $x = 2$,$h'(2) = f'(2)g(2) + g'(2)f(2) = (-2)(5) + (4)(3) = -10

Vedclass · Accepted Answer

$F'(2) = 22 h'(2)$

Vedclass · Answer

(B) Given $h(x) = f(x)g(x)$. Using the product rule,$h'(x) = f'(x)g(x) + g'(x)f(x)$.At $x = 2$,$h'(2) = f'(2)g(2) + g'(2)f(2) = (-2)(5) + (4)(3) = -10 + 12 = 2$.Given $F(x) = f(g(x))$. Using the chain rule,$F'(x) = f'(g(x)) \cdot g'(x)$.At $x = 2$,$F'(2) = f'(g(2)) \cdot g'(2) = f'(5) \cdot 4 = 11 \cdot 4 = 44$.Now,comparing $F'(2)$ and $h'(2)$,we have $F'(2) = 44$ and $h'(2) = 2$.Therefore,$F'(2) = 22 \cdot h'(2)$.

Suppose that $h(x) = f(x) \cdot g(x)$ and $F(x) = f(g(x))$,where $f(2) = 3$,$g(2) = 5$,$g'(2) = 4$,$f'(2) = -2$,and $f'(5) = 11$. Then:

Similar Questions

Assertion: For $x < 0$,$\frac{d^2}{d x^2}(\log |x|) = \frac{1}{|x|^2}$.
Reason: For $x < 0$,$|x| = -x$.

Let $f(x) = (x^x)^x$ and $g(x) = x^{(x^x)}$,then:

If $y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right)$,then $y'(1)$ is

If $y = \sqrt{\sin \sqrt{x}}$,then $\frac{dy}{dx} = $

If $y = \frac{1}{4}u^4$ and $u = \frac{2}{3}x^3 + 5$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module