Question

If the function $f\left( x \right)={{x}^{3}}+{{e}^{\dfrac{x}{2}}}$ and $g\left( x \right)={{f}^{-1}}\left( x \right)$, then the value of ${{g}^{'}}\left( 1 \right)$ is

Answer 1

Hint: For this problem we need to calculate the value of ${{g}^{'}}\left( 1 \right)$ where the given data is $f\left( x \right)={{x}^{3}}+{{e}^{\dfrac{x}{2}}}$ and $g\left( x \right)={{f}^{-1}}\left( x \right)$. We will first calculate the value of ${{f}^{'}}\left( x \right)$ by differentiating the given function $f\left( x \right)={{x}^{3}}+{{e}^{\dfrac{x}{2}}}$ with respect to $x$. Now from the second relation which is $g\left( x \right)={{f}^{-1}}\left( x \right)$, we will calculate the value of $g\left( f\left( x \right) \right)$ and differentiate it with respect to $x$. Now we can calculate the value of ${{g}^{'}}\left( 1 \right)$ by using the value $f\left( 0 \right)=1$.

Complete step by step solution:
Given that, $f\left( x \right)={{x}^{3}}+{{e}^{\dfrac{x}{2}}}$ and $g\left( x \right)={{f}^{-1}}\left( x \right)$.
Differentiating the given function $f\left( x \right)$ with respect to $x$, then we will get
${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ {{x}^{3}}+{{e}^{\dfrac{x}{2}}} \right]$
Applying differentiation for each term individually, then we will have
${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}} \right)+\dfrac{d}{dx}\left( {{e}^{\dfrac{x}{2}}} \right)$
Using the differentiation formulas $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, $\dfrac{d}{dx}\left( {{e}^{ax}} \right)=a{{e}^{ax}}$ in the above equation, then the above equation is modified as
${{f}^{'}}\left( x \right)=3{{x}^{2}}+\dfrac{1}{2}{{e}^{\dfrac{x}{2}}}$
Now consider the given equation $g\left( x \right)={{f}^{-1}}\left( x \right)$. From this equation we can write that
$g\left( f\left( x \right) \right)=x$
Differentiating the above equation with respect to $x$, then we will get
$\dfrac{d}{dx}\left[ g\left( f\left( x \right) \right) \right]=\dfrac{dx}{dx}$
Using the differentiation formula $\dfrac{d}{dx}\left[ p\left( q\left( x \right) \right) \right]={{p}^{'}}\left( q\left( x \right) \right).{{q}^{'}}\left( x \right)$ in the above equation, then the above equation is modified as
$\begin{align}
  & {{g}^{'}}\left( f\left( x \right) \right)\times {{f}^{'}}\left( x \right)=1 \\
 & \Rightarrow {{g}^{'}}\left( f\left( x \right) \right)=\dfrac{1}{{{f}^{'}}\left( x \right)} \\
\end{align}$
To find the value of ${{g}^{'}}\left( 1 \right)$, the value of $f\left( x \right)$ should be equal to $1$ in the above equation. By observing the given function $f\left( x \right)={{x}^{3}}+{{e}^{\dfrac{x}{2}}}$, we can write that $f\left( 0 \right)=1$. So, to find the value of ${{g}^{'}}\left( 1 \right)$ substituting $x=0$ in the above equation, then we will get
${{g}^{'}}\left( f\left( 0 \right) \right)=\dfrac{1}{{{f}^{'}}\left( 0 \right)}$
From the values $f\left( 0 \right)=1$, ${{f}^{'}}\left( x \right)=3{{x}^{2}}+\dfrac{1}{2}{{e}^{\dfrac{x}{2}}}$ the above equation is modified as
${{g}^{'}}\left( 1 \right)=\dfrac{1}{3{{\left( 0 \right)}^{2}}+\dfrac{1}{2}{{e}^{\dfrac{0}{2}}}}$
We know that the value of ${{e}^{\dfrac{0}{2}}}={{e}^{0}}=1$. Substituting this value in the above equation then we will get
$\begin{align}
  & {{g}^{'}}\left( 1 \right)=\dfrac{1}{\dfrac{1}{2}} \\
 & \therefore {{g}^{'}}\left( 1 \right)=2 \\
\end{align}$
Hence the value of the ${{g}^{'}}\left( 1 \right)$ is $2$.

Note: We can also solve this problem in another method which is the traditional method in which we will calculate the inverse of $f\left( x \right)$ and equate to the $g\left( x \right)$ and calculate the required value by differentiating the function $g\left( x \right)$ and substituting $x=1$ in ${{g}^{'}}\left( x \right)$. We can calculate the inverse of the function $f\left( x \right)$ by equating the given function to $y$ and calculating the value of $x$ in terms of $y$.