Let h(x) be differentiable for all x and let | vedclass

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

Question

(C) Given $f(x) = (kx + e^x) h(x)$.Using the product rule for differentiation,$f'(x) = \frac{d}{dx}(kx + e^x) \cdot h(x) + (kx + e^x) \cdot h'(x)$.$f'

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given $f(x) = (kx + e^x) h(x)$.Using the product rule for differentiation,$f'(x) = \frac{d}{dx}(kx + e^x) \cdot h(x) + (kx + e^x) \cdot h'(x)$.$f'(x) = (k + e^x) h(x) + (kx + e^x) h'(x)$.Now,substitute $x = 0$ into the derivative expression:$f'(0) = (k + e^0) h(0) + (k(0) + e^0) h'(0)$.$f'(0) = (k + 1) h(0) + (0 + 1) h'(0)$.$f'(0) = (k + 1) h(0) + h'(0)$.Given $h(0) = 5$,$h'(0) = -2$,and $f'(0) = 18$,we substitute these values:$18 = (k + 1)(5) + (-2)$.$18 = 5k + 5 - 2$.$18 = 5k + 3$.$15 = 5k$.$k = 3$.

Let $h(x)$ be differentiable for all $x$ and let $f(x) = (kx + e^x) h(x)$ where $k$ is some constant. If $h(0) = 5$,$h'(0) = -2$,and $f'(0) = 18$,then the value of $k$ is equal to

Similar Questions

If $y=a \sin x+b \cos x$,then $y^2+\left(\frac{d y}{d x}\right)^2$ is a

Let $f: R \rightarrow R$ be a function such that $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6, x \in R$,then $f(2)$ equals

If $A = \frac{2^x \cot x}{\sqrt{x}}$,then $\frac{dA}{dx} = $

$\frac{d}{dx}(\sin(x^2) + \cos(x^2)) = $ . . . . . .

If $f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+\tan x}$,then $f^{\prime}\left(\frac{\pi}{4}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module