Question

Find the derivative of ${{x}^{2}}\sin x$ from the first principle and verify from product rule method.

Answer 1

Consider the function ${{x}^{2}}\sin x$ as $f\left( x \right)$. Assume ‘h’ as the small change in x and hence find the small change in the function f (x) given as f(x + h). Now, use the limit definition of derivative and apply the formula $f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$, substitute the value of the given functions. Use the formulas $\left( \sin a-\sin b \right)=2\sin \left( \dfrac{a-b}{2} \right)\cos \left( \dfrac{a+b}{2} \right)$ and to simplify. Use the basic limit formula $\displaystyle \lim_{\theta \to 0}\dfrac{\sin \theta }{\theta }=1$ to get the answer. Now, to verify the answer using the rule method, consider the function ${{x}^{2}}$ as u and $\sin x$ as v, use the product rule of the derivative given as $\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. Use the formulas $\dfrac{d\left[ {{x}^{n}} \right]}{dx}=n{{x}^{n-1}}$ and $\dfrac{d\left( \sin x \right)}{dx}=\cos x$ to get the answer for the comparison.

Complete step-by-step answer:
Here we have been provided with the function ${{x}^{2}}\sin x$ and we are asked to find its derivative using the first principle. Also we need to verify the answer using the rule method which will be product rule of differentiation.
We know that derivative of a function is defined as the rate of change of function. In other words we can say that it is the measure of change in the value of the function with respect to the change in the value of the variable on which it depends. This change is infinitesimally small, that means tending to zero. Mathematically we have,
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right)$
In the above formula we have ‘h’ as the infinitesimally small change in the variable x.
Let us consider the function ${{x}^{2}}\sin x$ as $f\left( x \right)$, so substituting $\left( x+h \right)$ in place of x in the function we get,
$\Rightarrow f\left( x+h \right)={{\left( x+h \right)}^{2}}\sin \left( x+h \right)$
Now, substituting the value of the functions f (x) and f (x + h) in the limit formula we get,
$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{{{\left( x+h \right)}^{2}}\sin \left( x+h \right)-{{x}^{2}}\sin x}{h} \right)$
Using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ we get,
$\begin{aligned} & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{\left( {{x}^{2}}+{{h}^{2}}+2xh \right)\sin \left( x+h \right)-{{x}^{2}}\sin x}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{{{x}^{2}}\sin \left( x+h \right)+{{h}^{2}}\sin \left( x+h \right)+2xh\sin \left( x+h \right)-{{x}^{2}}\sin x}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{{{x}^{2}}\left( \sin \left( x+h \right)-\sin x \right)+{{h}^{2}}\sin \left( x+h \right)+2xh\sin \left( x+h \right)}{h} \right) \\\ \end{aligned}$
Breaking the terms we get,

& \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{{{x}^{2}}\left( \sin \left( x+h \right)-\sin x \right)}{h}+\dfrac{{{h}^{2}}\sin \left( x+h \right)}{h}+\dfrac{2xh\sin \left( x+h \right)}{h} \right) \\\ & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{{{x}^{2}}\left( \sin \left( x+h \right)-\sin x \right)}{h}+h\sin \left( x+h \right)+2x\sin \left( x+h \right) \right) \\\ \end{aligned}$$ Using the trigonometric identity $\left( \sin a-\sin b \right)=2\sin \left( \dfrac{a-b}{2} \right)\cos \left( \dfrac{a+b}{2} \right)$ we get, $$\begin{aligned} & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{{{x}^{2}}\left( 2\sin \left( \dfrac{x+h-x}{2} \right)\cos \left( \dfrac{x+h+x}{2} \right) \right)}{h}+h\sin \left( x+h \right)+2x\sin \left( x+h \right) \right) \\\ & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{2{{x}^{2}}\sin \left( \dfrac{h}{2} \right)\cos \left( \dfrac{2x+h}{2} \right)}{h}+h\sin \left( x+h \right)+2x\sin \left( x+h \right) \right) \\\ \end{aligned}$$ We can write simplify the first term of the limit and write the expression as: - $$\Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\left( \dfrac{{{x}^{2}}\sin \left( \dfrac{h}{2} \right)\cos \left( \dfrac{2x+h}{2} \right)}{\left( \dfrac{h}{2} \right)}+h\sin \left( x+h \right)+2x\sin \left( x+h \right) \right)$$ Separating the limits we can write the above expression as: - $$\Rightarrow f'\left( x \right)=\left( \displaystyle \lim_{h \to 0}\dfrac{{{x}^{2}}\sin \left( \dfrac{h}{2} \right)\cos \left( \dfrac{2x+h}{2} \right)}{\left( \dfrac{h}{2} \right)}+\displaystyle \lim_{h \to 0}h\sin \left( x+h \right)+\displaystyle \lim_{h \to 0}2x\sin \left( x+h \right) \right)$$ Using the formula $\displaystyle \lim_{\theta \to 0}\dfrac{\sin \theta }{\theta }=1$ for the first term and substituting the value of h = 0 in the other two terms we get, $$\begin{aligned} & \Rightarrow f'\left( x \right)=\left( \displaystyle \lim_{h \to 0}{{x}^{2}}\cos \left( \dfrac{2x+h}{2} \right)+0+2x\sin \left( x \right) \right) \\\ & \therefore f'\left( x \right)={{x}^{2}}\cos \left( x \right)+2x\sin \left( x \right) \\\ \end{aligned}$$ Therefore, the above relation is our answer. Now, let us verify the answer using the product rule of differentiation. Considering ${{x}^{2}}$ as u and $\sin x$ as v, using the formula $\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ we get, $\Rightarrow \dfrac{d\left( {{x}^{2}}\sin x \right)}{dx}={{x}^{2}}\dfrac{d\left( \sin x \right)}{dx}+\sin x\dfrac{d\left( {{x}^{2}} \right)}{dx}$ Using the basic formulas given as $\dfrac{d\left[ {{x}^{n}} \right]}{dx}=n{{x}^{n-1}}$ and $\dfrac{d\left( \sin x \right)}{dx}=\cos x$ we get, $\begin{aligned} & \Rightarrow \dfrac{d\left( {{x}^{2}}\sin x \right)}{dx}={{x}^{2}}\cos x+\sin x\left( 2x \right) \\\ & \Rightarrow \dfrac{d\left( {{x}^{2}}\sin x \right)}{dx}={{x}^{2}}\cos x+2x\sin x \\\ \end{aligned}$ Hence, we can see that the above answer is equal to the one we have obtained using the first principle so our answer is verified. **Note:** Note that all the formulas of derivatives of different functions are derived from the first principle. It is the basic definition of the derivative. Remember the basic rules of differentiation like the product rule, chain rule, $\dfrac{u}{v}$ rule etc. as they will help us to verify the answer or to solve the question in less steps. Also, remember the derivatives of common functions like the trigonometric, inverse trigonometric, logarithmic, exponential functions etc.