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Question: Differentiate \(x{{e}^{x}}\) from first principle....

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

Explanation

Solution

Hint: First, here we will consider the function f(x)=xexf\left( x \right)=x{{e}^{x}} . Then we will differentiate this using the formula of first principle i.e. given as f′(x)=lim⁡h→0 f(x+h)−f(x)hf'\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(x)=xexf\left( x \right)=x{{e}^{x}} . So, we will use the formula of first principle given as f′(x)=lim⁡h→0 f(x+h)−f(x)hf'\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(x+h)=(x+h)ex+hf\left( x+h \right)=\left( x+h \right){{e}^{x+h}} , f(x)=xexf\left( x \right)=x{{e}^{x}}
On substituting in the formula, we get
f′(x)=lim⁡h→0 (x+h)ex+h−xexhf'\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′(x)=lim⁡h→0 (xex+h+hex+h)−xexhf'\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 xexx{{e}^{x}} common, we get
f′(x)=lim⁡h→0 xex(eh−1)h+hex+hhf'\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′(x)=lim⁡h→0 (xex(eh−1)h+ex+h)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′(x)=xexlim⁡h→0 (eh−1)h+lim⁡h→0 ex+hf'\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′(x)=xex(e0−1)h+ex+0=xex+exf'\left( x \right)=x{{e}^{x}}\dfrac{\left( {{e}^{0}}-1 \right)}{h}+{{e}^{x+0}}=x{{e}^{x}}+{{e}^{x}}
f′(x)=ex(x+1)f'\left( x \right)={{e}^{x}}\left( x+1 \right)
Thus, the answer of function after differentiation using the first principle is f′(x)=ex(x+1)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′(x)=lim⁡h→0 (xex+h+hex+h)−xexhf'\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′(x)=lim⁡h→0 xex+hh+lim⁡h→0 hex+hh−lim⁡h→0 xexhf'\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′(x)=xex+00+ex+0−xex0=∞+ex−∞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.