# Derivative of e^x^2 by First Principle and Chain Rule

The function e to the power x2 is written as $e^{x^2}$ and its derivative is $2xe^{x^2}$. In this post, we will find the derivative of e to the power x square by the first principle and chain rule of derivatives.

Recall the first principle of derivatives: The derivative of a function f(x) by first principle is defined by the limit:

$\dfrac{d}{dx}(f(x))$ = limh→0 $\dfrac{f(x+h)-f(x)}{h}$ $\cdots$ (I)

We will use this to find the differentiation of e to the power x2.

## Derivative of $e^{x^2}$ by First Principle

Let us put $f(x)=e^{x^2}$ in the above formula (I).

So we obtain that

$\dfrac{d}{dx}(e^{x^2})$ = limh→0 $\dfrac{e^{(x+h)^2}-e^{x^2}}{h}$

= limh→0 $\dfrac{e^{x^2+2xh+h^2}-e^{x^2}}{h}$. Here we have used the formula (a+b)2 = a2+2ab+b2.

= limh→0 $\dfrac{e^{x^2}e^{2xh+h^2}-e^{x^2}}{h}$

= limh→0 $\dfrac{e^{x^2}(e^{2xh+h^2}-1)}{h}$

= limh→0 $\dfrac{e^{x^2}(e^{h(2x+h)}-1)}{h}$

= limh→0 $[\dfrac{e^{x^2}(e^{h(2x+h)}-1)}{h(2x+h)}$ $\times (2x+h) ]$

$=e^{x^2}\lim\limits_{h \to 0}$ $\dfrac{e^{h(2x+h)}-1}{h(2x+h)}$ $\times \lim\limits_{h \to 0} (2x+h)$

Let u=h(2x+h). Then u tends to zero when h tends to 0.

$=e^{x^2}\lim\limits_{u \to 0}$ $\dfrac{e^{u}-1}{u}$ $\times \lim\limits_{h \to 0} (2x+h)$

$=e^{x^2} \times 1$ $\times (2x+0)$ as the limit of (ex-1)/x is 1 when x tends to 0.

$=2xe^{x^2}$

Thus, the derivative of e^x^2 by first principle is 2xe^x^2.

## Derivative of $e^{x^2}$ by Chain Rule

Let u=x2.

Then $\dfrac{du}{dx}=2x$.

Now, the derivative of $e^{x^2}$ by the chain rule is equal to

$\dfrac{d}{dx}(e^{x^2})$ $=\dfrac{d}{dx}(e^u)$

$= \dfrac{d}{du}(e^u) \cdot \dfrac{du}{dx}$

$=e^u \cdot 2x$ as du/dx=2x.

$=2xe^{x^2}$ as u=x2.

Therefore, the differentiation of e to the power x2 is equal to the product of 2x and e^{x^2}.

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## FAQs

Q1: What is the derivative of e^x^2?

Answer: The derivative of $e^{x^2}$ is equal to $2x e^{x^2}$.