Question

How do you differentiate \[f(x)={{x}^{2}}\cos x\] using product rule?

Answer 1

Hint: The product rule is a rule for differentiating expressions in which one function is multiplied by another function. The rule follows from the limit definition of derivative
and is given by \[\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] if \[y=uv\] where \[\dfrac{dy}{dx}\] means derivative of y with respect to x, \[\dfrac{dv}{dx}\] means derivative of v with respect to x and \[\dfrac{du}{dx}\] means derivative of u with respect to x. The above formula is applicable only if u and v are differentiable.

Complete step by step answer:
As per the given question, we have to differentiate the given function using the product rule. Here, the given function to be differentiated is \[f(x)={{x}^{2}}\cos x\].
Now, let \[y=f(x)\] then \[u={{x}^{2}}\] and \[v=\cos x\]. We know that, according to power rule,
derivative of \[a{{x}^{n}}\] is given by \[\dfrac{d}{dx}(a{{x}^{n}})=(na){{x}^{n-1}}\] and the derivative of trigonometric function \[\cos x\] is \[\dfrac{d}{dx}(\cos x)=-\sin x\].
The derivative of function u: -
By comparing the function ‘u’ with the function of power rule, we get the coefficient \[a=2\] and power of x, \[n=1\]. Then,
\[\Rightarrow \dfrac{du}{dx}=2x\] (\[\Leftarrow \]according to power rule)
The derivative of function v: -
By comparing the function ‘v’ with the trigonometric function, function ‘v’ is the same as the above trigonometric function. Then,
  \[\Rightarrow \]\[\dfrac{d}{dx}(\cos x)=-\sin x\] (\[\Leftarrow \]from derivative of logarithmic function)
Now substituting the above derivatives in product rule, we get
 \[\Rightarrow \]\[\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}={{x}^{2}}\dfrac{d(\cos x)}{dx}+\cos x\dfrac{d({{x}^{2}})}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}={{x}^{2}}(-\sin x)+\cos x(2x) \\
\end{align}\]
On solving the above equation, we get
\[\Rightarrow \dfrac{dy}{dx}=2x\cos x-{{x}^{2}}\sin x=x(2\cos x-x\sin x)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d(f(x)}{dx}=x(2\cos x-x\sin x)\]

\[\therefore \] The derivative of the given function \[f(x)={{x}^{2}}\cos x\] is \[x(2\cos x-x\sin x)\].

Note: In order to solve these types of problems, we must have enough knowledge about derivatives of basic functions. The product rule is nothing but take the derivative of v multiplied by u and add v multiplied by the derivative of u. We have the product rule for 3 functions also which is given by
\[\dfrac{dy}{dx}=\dfrac{df(x)}{dx}g(x)h(x)+f(x)\dfrac{dg\left( x \right)}{dx}h(x)+f(x)g(x)\dfrac{dh(x)}{dx}\] if \[y=f(x)g(x)h(x)\].