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

& \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} \\\ & y={{e}^{{{x}^{2}}}} \\\ \end{aligned}$$ Let us put, $$y={{e}^{u}}$$, where $$u={{x}^{2}}$$. $$\therefore \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$$ $$\begin{aligned} & \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{aligned}$$ We know, $$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.