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

Compute the derivative of $f(x) = \sin 2x$.

Find the derivative of $x^{5}(3-6x^{-9}).$

If $f'(x)$ is zero in the interval $(a, b)$,then in this interval it is

Find the derivative of $2x - \frac{3}{4}$.

If $f(x) = x \tan^{-1} x$,then $\lim_{x \rightarrow 1} \frac{f(x) - f(1)}{x - 1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module