Question

What is the derivative of \[{e^{2{x^2}}}\]?

Answer 1

Hint: Here, the given question has a trigonometric function. We have to find the derivative or differentiated term of the function. First consider the function \[y\], then differentiate \[y\] with respect to \[x\] by using a standard differentiation formula of trigonometric ratio and use chain rule for differentiation. And on further simplification we get the required differentiate value.

Complete step-by-step answer:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Consider the given function
\[y = {e^{2{x^2}}} - - - - \left( 1 \right)\]
We know the chain rule, that is \[y = {e^{g(x)}}\] then the derivative is given by
\[{y^1} = {e^{g(x)}}g'\left( x \right)\].
Here \[g\left( x \right) = 2{x^2}\].
now differentiating (1) with respect to x
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^{2{x^2}}}} \right)\]
\[\dfrac{{dy}}{{dx}} = {e^{2{x^2}}}\dfrac{d}{{dx}}\left( {2{x^2}} \right)\].
We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\].
\[\dfrac{{dy}}{{dx}} = {e^{2{x^2}}}\left( {2.2{x^{2 - 1}}} \right)\]
\[\dfrac{{dy}}{{dx}} = {e^{2{x^2}}}\left( {4x} \right)\]
\[\dfrac{{dy}}{{dx}} = 4x{e^{2{x^2}}}\]
Thus the derivative of \[{e^{2{x^2}}}\] with respect to x is \[4x{e^{2{x^2}}}\].
So, the correct answer is “ \[4x{e^{2{x^2}}}\]”.

Note: We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\]. The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero