Question

If \[y={{e}^{{{x}^{2}}}}\], then what is \[\dfrac{dy}{dt}\] at \[x=\pi \] equal to :
a.\[\left( 1+\pi \right){{e}^{{{\pi }^{2}}}}\]
b.\[2\pi {{e}^{{{\pi }^{2}}}}\]
c.\[2{{e}^{{{\pi }^{2}}}}\]
d.\[{{e}^{{{\pi }^{2}}}}\]

Answer 1

Hint: Using the chain rule find the derivative of the composite function. Put, \[u={{x}^{2}}\], then differentiate \[\dfrac{dy}{du}\] and \[\dfrac{du}{dx}\] multiply it together to get \[\dfrac{dy}{dx}\].

Complete step-by-step answer:

We have been given that, \[y={{e}^{{{x}^{2}}}}\].
We can differentiate the given function using chain rule.
The rule applied for finding the derivative of composition of function is basically known as the chain rule.
Let y represent a real valued function which is composite of two functions g and f such that :
\[y=f\left( g\left( x \right) \right)\], where we have found \[\dfrac{dy}{dx}\].
We need to substitute, \[u=g\left( x \right)\], which gives us \[y=f\left( u \right)\].
Then we need to use the formula of chain rule, which is
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} \\
 & y={{e}^{{{x}^{2}}}} \\
\end{align}\]
Let us put, \[y={{e}^{u}}\], where \[u={{x}^{2}}\].
\[\therefore \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}\]
\[\begin{align}
  & \therefore \dfrac{dy}{du}=\dfrac{d}{du}{{e}^{u}}={{e}^{u}} \\
 & \dfrac{du}{dx}=\dfrac{d}{dx}{{x}^{2}}=2x \\
 & \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}={{e}^{u}}\times \left( 2x \right)=2x{{e}^{u}} \\
\end{align}\]
We know, \[u={{x}^{2}}\].
\[\therefore \dfrac{dy}{dx}=2x{{e}^{{{x}^{2}}}}\]
Now it is said that \[x=\pi \], so substitute this value in \[\dfrac{dy}{dx}\].
\[\dfrac{dy}{dx}=2\pi {{e}^{{{\pi }^{2}}}}\]
Thus we got the required value, \[\dfrac{dy}{dx}=2\pi {{e}^{{{\pi }^{2}}}}\].
\[\therefore \] Option (b) is the correct answer.

Note: As the name of chain rule itself suggests that chain rule means differentiating the term one by one in a chain form starting from the outermost function to the innermost function.