Question

Find the derivative of \[x(\sin x)\] using the method of first principle.

Answer 1

Hint: First principle is a method of finding the derivative of any function, say f(x). It can be written as \[{{f}^{'}}(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\] (where \[{{f}^{'}}(x)\] refers to the derivative of the function \[f(x)\] ).

Complete step-by-step answer:

As mentioned in the question given above, the function of which the derivative is to be found by using the method of first principle is given as \[x(\sin x)\] .
Hence, by applying the formula for finding the derivative of the above function by the method of first principle, we get
\[\begin{align}
  & f(x)=x\sin x \\
 & {{f}^{'}}(x)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{(x+h)\sin (x+h)-x\sin x}{h} \right) \\
 & \ \ \ \ \ \ \ \ \ =\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{x\sin (x+h)+h\sin (x+h)-x\sin x}{h} \right) \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{x(\sin (x+h)-\sin (x))+h\sin (x+h)}{h} \right) \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{x\left( 2\cos \left( \dfrac{x+h+x}{2} \right)\sin \left( \dfrac{x+h-x}{2} \right) \right)+h\sin (x+h)}{h} \right) \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{x\left( 2\cos \left( \dfrac{2x+h}{2} \right)\sin \left( \dfrac{h}{2} \right) \right)+h\sin (x+h)}{h} \right) \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{x\left( 2\cos \left( \dfrac{2x+h}{2} \right)\sin \left( \dfrac{h}{2} \right) \right)}{h}+\dfrac{h\sin (x+h)}{h} \right) \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{x\left( \cos \left( \dfrac{2x+h}{2} \right)\sin \left( \dfrac{h}{2} \right) \right)}{\dfrac{h}{2}}+\sin (x+h) \right)\ \ \ \ \ ...(1) \\
 & =\left( x\left( \cos \left( \dfrac{2x}{2} \right) \right)+\sin (x) \right) \\
 & =\left( x\left( \cos \left( x \right) \right)+\sin (x) \right) \\
 & =x\cos x+\sin x \\
\end{align}\]
We used \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1\] in equation (1) in the above solution.
Hence, the derivative of xsinx comes out to be xcosx+sinx.

Note: Another alternative way of finding out the derivative of the above mentioned function is by using the product rule that is derivative of function uv is equal to
\[(uv)'=(u')v+u(v')\]
As it was mentioned in the question that we have to perform and find out the derivative of the given function using the method of first principle therefore we proceeded with that method. Otherwise one should always try to refrain from performing the method of first principle as it is lengthy and requires a lot of effort to find the derivative by using the method of first principle.