Question

Find the differential coefficient of the function ${{x}^{2}}.\log x.\sin x$ with respect to $x$ .

Answer 1

Hint : Differential coefficient of a function is nothing but the derivative of that function with respect to a given variable. So in order to solve this question we have to find the derivative of a given function with respect to $x$ . In order to find the derivative of the given function we have to apply product rule of the derivative, given by ${{F}^{'}}(x)=\dfrac{d}{dx}(f(x)).g(x)+f(x).\dfrac{d}{dx}(g(x))$ , where F(x)=f(x).g(x)

Complete step-by-step answer :
Let us consider that $h(x)={{x}^{2}}.\log x.\sin x$
So, the differential coefficient of the function ${{x}^{2}}.\log x.\sin x$ with respect to $x$ is equal to getting derivative of $h(x)$ w.r.t to $x$ .
Here, $h(x)$ is a multiplication of three functions ${{x}^{2}},\log x$ and $\sin x$ . So, whenever more than one function is multiplied with each other we use product rule to get the derivative.
According to product rule of derivative,
If $F(x)=f(x).g(x)$
Then, ${{F}^{'}}(x)=\dfrac{d}{dx}(f(x)).g(x)+f(x).\dfrac{d}{dx}(g(x))$
Similarly this can be extended for the product of three functions too, which is
If $F(x)=f(x).g(x).m(x)$
Then, ${{F}^{'}}(x)=\dfrac{d}{dx}(f(x)).g(x).m(x)+f(x).\dfrac{d}{dx}(g(x)).m(x)+f(x).g(x).\dfrac{d}{dx}(m(x))$
So now, we have $h(x)={{x}^{2}}.\log x.\sin x$
Applying product rule, we get
$\Rightarrow {{h}^{'}}(x)=\dfrac{d}{dx}({{x}^{2}}).\log x.\sin x+{{x}^{2}}.\dfrac{d}{dx}(\log x).\sin x+{{x}^{2}}.\log x.\dfrac{d}{dx}(\sin x)\ldots \ldots \text{ }\left( 1 \right)$
Now, we know that the derivative of ${{x}^{n}}$ is $n{{x}^{n-1}}$ , as
$\therefore \dfrac{d}{dx}({{x}^{n}})=n.{{x}^{n-1}}\Rightarrow \dfrac{d}{dx}({{x}^{2}})=2.{{x}^{2-1}}=2x$
We also know that derivative of $\log x$ and $\sin x$ is $\dfrac{1}{x}$ and $\cos x$ respectively, as
$\therefore \dfrac{d}{dx}(\log x)=\dfrac{1}{x}$ and $\dfrac{d}{dx}(\sin x)=\cos x$
Putting all these values in equation (1), we get
$\Rightarrow {{h}^{'}}(x)=2x.\log x.\sin x+{{x}^{2}}.\dfrac{1}{x}.\sin x+{{x}^{2}}.\log x.\cos x$
After simplifying the above equation, we get
$\Rightarrow {{h}^{'}}(x)=2x.\log x.\sin x+x.\sin x+{{x}^{2}}.\log x.\cos x$
And so the differential coefficient of the function ${{x}^{2}}.\log x.\sin x$ with respect to $x$ is $2x.\log x.\sin x+x.\sin x+{{x}^{2}}.\log x.\cos x$
Hence, the answer is $2x.\log x.\sin x+x.\sin x+{{x}^{2}}.\log x.\cos x$ .

Note : This is a pretty straight forward question as there is nothing tricky in this question. One thing that students find confusing is the differential coefficient of the function which is nothing but the derivative of the function. After figuring this thing out then we have to find the derivative of the function in order to get the derivative of the function.