Differentiate $x{e^{{x^2}}}$ .

Hint: Differentiation is a process to find the rate of change of a function with respect to one of it’s variables. If $x$ is a variable and $y$ represents another variable, then the rate of change of $y$ w.r.t. $x$ is given as $\dfrac{{dy}}{{dx}}$ . The general expression for the derivative of a function is represented by, $f'\left( x \right) = \dfrac{{dy}}{{dx}},{\text{ where }}f\left( x \right){\text{ is any function}}{\text{.}}$ To solve this question , we should be familiar with some standard derivatives and fundamental rules of differentiation.

Complete step by step answer:
Let $y = x{e^{{x^2}}}{\text{ }}......\left( 1 \right)$
Differentiate the above equation w.r.t. $x$ , we get;
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {x \times {e^{{x^2}}}} \right)$
By the product rule of differentiation, we know that;
$ \Rightarrow \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}{\text{ }}......\left( 2 \right)$
Here , $u = x{\text{ and }}v = {e^{{x^2}}}$
Now, let us calculate $\dfrac{{du}}{{dx}}$ .
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( x \right)$
$ \Rightarrow \dfrac{{du}}{{dx}} = 1{\text{ }}......\left( 3 \right)$
Similarly, the value of $\dfrac{{dv}}{{dx}}$ will be;
$ \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^{{x^2}}}} \right)$
By chain rule of differentiation, which is used to calculate the derivative of a composite function or we can say a function within a function. For example: $\sin \left( {{x^2}} \right)$ is a composite function. So first we will calculate the derivative of $\sin x$ w.r.t. $x$ and then as we can notice that ${x^2}$ is also differentiable w.r.t. $x$ . Therefore, we will also calculate it’s derivative which will be $2x$ . So, $\dfrac{d}{{dx}}\sin \left( {\left( {{x^2}} \right)} \right) = 2x\cos x$.
We know that $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ and also applying the chain rule of differentiation here;
$ \Rightarrow \dfrac{{dv}}{{dx}} = {e^{{x^2}}} \times 2x{\text{ }}......\left( 4 \right)$
Now put the values of $\dfrac{{du}}{{dx}}{\text{ and }}\dfrac{{dv}}{{dx}}{\text{ }}$ from equation $\left( 3 \right){\text{ and }}\left( 4 \right)$ in equation $\left( 2 \right)$, we get;
$ \Rightarrow \dfrac{d}{{dx}}\left( {uv} \right) = \left( {{e^{{x^2}}} \times 1} \right) + \left( {x \times {e^{{x^2}}} \times 2x} \right){\text{ }}$
Simplifying further, we get;
$ \Rightarrow \dfrac{d}{{dx}}\left( {uv} \right) = {e^{{x^2}}}{\text{ + }}\left( {{\text{2}}{x^2} \times {e^{{x^2}}}} \right){\text{ }}$
Taking ${e^{{x^2}}}$ outside the equation, we get;
$ \Rightarrow \dfrac{d}{{dx}}\left( {uv} \right) = {e^{{x^2}}}\left( {1 + 2{x^2}} \right){\text{ }}$
Therefore, $\dfrac{{dy}}{{dx}} = {e^{{x^2}}}\left( {1 + 2{x^2}} \right)$
Therefore the correct answer for this question is the differentiation of $x{e^{{x^2}}}{\text{ is }}{{\text{e}}^{{x^2}}}\left( {1 + 2{x^2}} \right)$ .
The standard derivatives of some important functions are given below;
$\left( 1 \right)\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ . $\left( 2 \right)\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ . $\left( 3 \right)\dfrac{d}{{dx}}\left( {{{\log }_e}x} \right) = \dfrac{1}{x}$ . $\left( 4 \right)\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$. . $\left( 5 \right)\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$ . $\left( 6 \right)\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$ $\left( 7 \right)\dfrac{d}{{dx}}\left( {\cot x} \right) = - \cos e{c^2}x$ . . $\left( 8 \right)\dfrac{d}{{dx}}\left( {\cos ecx} \right) = - \cos ecx\tan x$ .

Note: There are some fundamental rules for differentiation, which are stated as follows: $\left( 1 \right)$ Differentiation of a constant is always zero, i.e. $\dfrac{d}{{dx}}\left( {{\text{constant}}} \right) = 0$ . $\left( 2 \right)$ Differentiation of a constant multiplied with a function is calculated as constant times differentiation of that function. For example: $\dfrac{d}{{dx}}\left( {3\sin x} \right) = 3\dfrac{d}{{dx}}\left( {\sin x} \right)$ which gives the result as $3\cos x$ . $\left( 3 \right)$The product rule of differentiation is given as $\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) \times g\left( x \right) + f\left( x \right) \times \dfrac{d}{{dx}}g\left( x \right)$ . $\left( 4 \right)$ The division rule of differentiation is given by $\dfrac{d}{{dx}}\left\{ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \dfrac{{g\left( x \right) \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) - f\left( x \right) \times \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$ .