Question

Differentiate $x{{e}^{x}}$ from first principle.

Answer 1

Hint: First, here we will consider the function $f\left( x \right)=x{{e}^{x}}$ . Then we will differentiate this using the formula of first principle i.e. given as $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . Then, we will substitute the value of limit at the end. Thus, we will get the required answer.

Complete step-by-step solution -
First principle is also known as the delta method. Now, we will consider the function $f\left( x \right)=x{{e}^{x}}$ . So, we will use the formula of first principle given as $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . So, finding the value we get,
$f\left( x+h \right)=\left( x+h \right){{e}^{x+h}}$ , $f\left( x \right)=x{{e}^{x}}$
On substituting in the formula, we get
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( x+h \right){{e}^{x+h}}-x{{e}^{x}}}{h}$
On multiplying the brackets, we get
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( x{{e}^{x+h}}+h{{e}^{x+h}} \right)-x{{e}^{x}}}{h}$
On taking $x{{e}^{x}}$ common, we get
\[f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,x{{e}^{x}}\dfrac{\left( {{e}^{h}}-1 \right)}{h}+\dfrac{h{{e}^{x+h}}}{h}\]
On further simplification, we get
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left(x{{e}^{x}}\dfrac{\left( {{e}^{h}}-1 \right)}{h}+{{e}^{x+h}}\right)$
Now taking limit individually, we get
$f'\left( x \right)=x{{e}^{x}}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( {{e}^{h}}-1 \right)}{h}+\underset{h\to 0}{\mathop{\lim }}\,{{e}^{x+h}}$
On putting the value of h as zero, we get the bracket term as 1. So, we get
$f'\left( x \right)=x{{e}^{x}}\dfrac{\left( {{e}^{0}}-1 \right)}{h}+{{e}^{x+0}}=x{{e}^{x}}+{{e}^{x}}$
$f'\left( x \right)={{e}^{x}}\left( x+1 \right)$
Thus, the answer of function after differentiation using the first principle is $f'\left( x \right)={{e}^{x}}\left( x+1 \right)$ .

Note: Be careful while doing limit problems. It is very tricky to understand when to put a value of limit. Not knowing this can change the whole answer. Suppose from the above solution we take $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( x{{e}^{x+h}}+h{{e}^{x+h}} \right)-x{{e}^{x}}}{h}$ as $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{x{{e}^{x+h}}}{h}+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{h{{e}^{x+h}}}{h}-\underset{h\to 0}{\mathop{\lim }}\,\dfrac{x{{e}^{x}}}{h}$ and putting value of h as 0, we get answer as $f'\left( x \right)=\dfrac{x{{e}^{x+0}}}{0}+{{e}^{x+0}}-\dfrac{x{{e}^{x}}}{0}=\infty +{{e}^{x}}-\infty $ . So, we cannot write any specific answer here and this is the wrong answer. So, please be careful while putting the value of limit.