If f(x)=e^{x} g(x),g(0)=4,and g^{prime}(0)=2, | vedclass

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

Question

(B) Given the function $f(x) = e^{x} g(x)$.Applying the product rule for differentiation,we have $f^{\prime}(x) = \frac{d}{dx}(e^{x}) \cdot g(x) + e^{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given the function $f(x) = e^{x} g(x)$.Applying the product rule for differentiation,we have $f^{\prime}(x) = \frac{d}{dx}(e^{x}) \cdot g(x) + e^{x} \cdot \frac{d}{dx}(g(x))$.This simplifies to $f^{\prime}(x) = e^{x} g(x) + e^{x} g^{\prime}(x) = e^{x} (g(x) + g^{\prime}(x))$.Now,substitute $x = 0$ into the derivative expression:$f^{\prime}(0) = e^{0} (g(0) + g^{\prime}(0))$.Given $g(0) = 4$ and $g^{\prime}(0) = 2$,and knowing $e^{0} = 1$:$f^{\prime}(0) = 1 \cdot (4 + 2) = 6$.

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

Similar Questions

If $f^{\prime}(x) = \tan^{-1}(\sec x + \tan x)$,$\frac{-\pi}{2} < x < \frac{\pi}{2}$ and $f(0) = 0$,then $f(1) =$

$A$ function $f: R \rightarrow R$ is such that $y f(x+y) + \cos(mxy) = 1 + y f(x)$. If $m=2$, then $f'(x) =$

The value of $\Delta \log f(x) + \Delta^{2}(3^{x})$ is

Find the derivative: $\frac{d}{dx} \left( \frac{1}{x^4 \sec x} \right) = $

Find the derivative of the following function: $5 \sec x + 4 \cos x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module