(B) Given $f(x) = e^{x} g(x)$.Applying the product rule for differentiation,we get:$f^{\prime}(x) = e^{x} g^{\prime}(x) + e^{x} g(x)$.Substituting $x

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(B) Given $f(x) = e^{x} g(x)$.Applying the product rule for differentiation,we get:$f^{\prime}(x) = e^{x} g^{\prime}(x) + e^{x} g(x)$.Substituting $x = 0$:$f^{\prime}(0) = e^{0} \cdot g^{\prime}(0) + e^{0} \cdot g(0)$.Since $e^{0} = 1$,$g^{\prime}(0) = 1$,and $g(0) = 2$:$f^{\prime}(0) = 1 \cdot 1 + 1 \cdot 2$.$f^{\prime}(0) = 1 + 2 = 3$.

Class 12 Mathematics Continuity and Differentiability Derivative at a point, Standard differentiation

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If $f(x) = e^{x} g(x)$,$g(0) = 2$,and $g^{\prime}(0) = 1$,then $f^{\prime}(0)$ is

MHT CET 2011 All MHT CET

A
$1$
B
$3$
C
$2$
D
$0$

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