Question

If \[f\left( x \right) = {e^x}g\left( x \right)\], \[g\left( 0 \right) = 2\], and \[g'\left( 0 \right) = 1\]. Then what is the value of \[f'\left( 0 \right)\]?
A. 1
B. 2
C. 3
D. 0

Answer 1

Hint: First, differentiate the given function with respect to \[x\]. Then substitute \[x = 0\] in the differential equation. In the end, substitute the given values in the differential equation and get the required answer.

Formula Used:
\[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\]
Product rule of differentiation: \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]

Complete step by step solution:
The given function is \[f\left( x \right) = {e^x}g\left( x \right)\].
\[g\left( 0 \right) = 2\], and \[g'\left( 0 \right) = 1\]
Let’s differentiate the given function with respect to \[x\].
\[\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {{e^x}g\left( x \right)} \right)\]
Apply the product rule of differentiation on the right-hand side.
\[f'\left( x \right) = {e^x}\dfrac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)\]
\[ \Rightarrow \]\[f'\left( x \right) = {e^x}g'\left( x \right) + g\left( x \right)\left( {{e^x}} \right)\] [ Since \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\]]
Factor out the common term.
\[f'\left( x \right) = {e^x}\left[ {g'\left( x \right) + g\left( x \right)} \right]\]

Now substitute \[x = 0\] in the above function.
\[f'\left( 0 \right) = {e^0}\left[ {g'\left( 0 \right) + g\left( 0 \right)} \right]\]
\[ \Rightarrow \]\[f'\left( 0 \right) = 1\left[ {g'\left( 0 \right) + g\left( 0 \right)} \right]\] [ Since \[{e^0} = 1\]]
Substitute the given values in the above equation.
\[f'\left( 0 \right) = 1 + 2\]
\[ \Rightarrow \]\[f'\left( 0 \right) = 3\]

Hence the correct option is C.

Note: The derivative is a rate of change of a function with respect to a variable.
The \[{n^{th}}\] derivative of an exponential function \[{e^x}\] is: \[\dfrac{{{d^n}{e^x}}}{{d{x^n}}} = {e^x}\]
The derivative of an exponential function \[{e^{ax}}\] is: \[\dfrac{d}{{dx}}\left( {{e^{ax}}} \right) = a{e^{ax}}\]