Question

If \[f\left( x \right)=x\cos x\], then find \[{f}''\left( x \right)\].

Answer 1

Hint: We solve this problem simply by using the product rule of differentiation. The product rule of differentiation is given as if \[u,v\] are two functions of \['x'\] then,
\[\dfrac{d}{dx}\left( u.v \right)=u.\dfrac{d}{dx}\left( v \right)+v.\dfrac{d}{dx}\left( u \right)\]
After applying the above rule we use the standard results of differentiation as
\[\begin{align}
  & \dfrac{d}{dx}\left( \cos x \right)=-\sin x \\
 & \dfrac{d}{dx}\left( \sin x \right)=\cos x \\
\end{align}\]
By using these formulas we calculate the value of \[{f}''\left( x \right)\].

Complete step by step answer:
We are given with a function as
\[f\left( x \right)=x\cos x\]
We are asked to find the double differentiation of the above function.
Let us find the first derivative of the given function.
By applying the differential operator on both sides we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( x\cos x \right) \\
 & \Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( x\cos x \right)..........equation\left( i \right) \\
\end{align}\]
We know that the product rule of differentiation is given as if\[u,v\] are two functions of \['x'\] then,
\[\dfrac{d}{dx}\left( u.v \right)=u.\dfrac{d}{dx}\left( v \right)+v.\dfrac{d}{dx}\left( u \right)\]
Let us assume the functions \[u,v\] as
\[\begin{align}
  & \Rightarrow u=x \\
 & \Rightarrow v=\cos x \\
\end{align}\]
Now, by using the product rule of differentiation to equation (i) we get
\[\Rightarrow {f}'\left( x \right)=x.\dfrac{d}{dx}\left( \cos x \right)+\cos x.\dfrac{d}{dx}\left( x \right)\]
We know that the standard results of differentiation as
\[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\]
By substituting the standard result in above equation we get
\[\begin{align}
  & \Rightarrow {f}'\left( x \right)=x\left( -\sin x \right)+\cos x\left( 1 \right) \\
 & \Rightarrow {f}'\left( x \right)=-x\sin x+\cos x \\
\end{align}\]
We are asked to find the double derivative.
Now, by applying the differential operator on both sides to above equation we get
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( {f}'\left( x \right) \right)=\dfrac{d}{dx}\left( -x\sin x \right)+\dfrac{d}{dx}\left( \cos x \right) \\
 & \Rightarrow {f}''\left( x \right)=\dfrac{d}{dx}\left( -x\sin x \right)+\dfrac{d}{dx}\left( \cos x \right) \\
\end{align}\]
Similarly, by using the product rule to above equation we get
\[\Rightarrow {f}''\left( x \right)=-x\dfrac{d}{dx}\left( \sin x \right)+\sin x\dfrac{d}{dx}\left( -x \right)+\dfrac{d}{dx}\left( \cos x \right)....equation(ii)\]
We know the standard results of differentiation as
\[\begin{align}
  & \dfrac{d}{dx}\left( \cos x \right)=-\sin x \\
 & \dfrac{d}{dx}\left( \sin x \right)=\cos x \\
\end{align}\]
By using these standard results to equation (ii) we get
\[\begin{align}
  & \Rightarrow {f}''\left( x \right)=-x\left( \cos x \right)+\sin x\left( -1 \right)+\left( -\sin x \right) \\
 & \Rightarrow {f}''\left( x \right)=-x\cos x-2\sin x \\
\end{align}\]

Therefore, we can say that the double derivative of a given function is \[{f}''\left( x \right)=-x\cos x-2\sin x\].

Note: Students may make mistakes at the point of second derivative. We are asked to find \[{f}''\left( x \right)\]. But, students may stop the solution at the following point
\[\Rightarrow {f}'\left( x \right)=-x\sin x+\cos x\]
This is because of over reading. Students may find only the first derivative due to over reading and do not find the second derivative. Also students may take the standard values of differentiation wrong. The standard results of differentiation used in the question are
\[\begin{align}
  & \dfrac{d}{dx}\left( \cos x \right)=-\sin x \\
 & \dfrac{d}{dx}\left( \sin x \right)=\cos x \\
\end{align}\]
Students may do mistake here taking
\[\dfrac{d}{dx}\left( \cos x \right)=\sin x\]
This results in a wrong answer. This part needs to be taken care of.