Question

Differentiate the function with respect to x using first principal

$f(x) = {xe}^{x}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = x \: {e}^{x}$

So,

$\rm \: f(x + h) = (x + h) \: {e}^{x + h}$

By using definition of First Principal, we have

$\boxed{\sf{ \: \: f'(x) \: = \: \displaystyle\lim_{h \to 0}\sf \: \frac{f(x + h) - f(x)}{h} \: \: }}$

So, on substituting the values, we get

$\rm \: f'(x) = \displaystyle\lim_{h \to 0}\rm \frac{(x + h){e}^{x + h} - x{e}^{x}}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{x{e}^{x + h} + h{e}^{x + h} - x{e}^{x}}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{x({e}^{x + h} - {e}^{x}) + h{e}^{x + h}}{h}$

can be further rewritten as

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{x({e}^{x}{e}^{h} - {e}^{x})}{h} + \displaystyle\lim_{h \to 0}\rm \frac{h{e}^{x + h}}{h}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{x{e}^{x}({e}^{h} - 1)}{h} + \displaystyle\lim_{h \to 0}\rm {e}^{x + h}$

We know,

$\boxed{\sf{ \: \: \displaystyle\lim_{x \to 0}\rm \frac{{e}^{x} - 1}{x} = 1 \: }}$

So, using this identity, we get

$\rm \: = \: x{e}^{x} \times 1 + {e}^{x + 0}$

$\rm \: = \: x{e}^{x} + {e}^{x}$

$\rm \: = \: {e}^{x}(x + 1)$

Hence,

$\boxed{\sf{ \:\: \: \rm \: \dfrac{d}{dx}x \: {e}^{x} = \: {e}^{x}(x + 1) \: \: }}$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$