Question

How do you find the second derivative of $y=\cos \left( {{x}^{2}} \right)$ ?

Answer 1

To solve the given equation, we need to get the first derivative by using Chain Rule,
\dfrac{d\left[ f\left\\{ g\left( x \right) \right\\} \right]}{dx}=f'\left\\{ g\left( x \right) \right\\}g'\left( x \right) .
Then to get a second derivative we need to use the Product Rule as well as the Chain Rule.
Product Rule - $\dfrac{d\left[ f\left( x \right)g\left( x \right) \right]}{dx}=f\left( x \right)\dfrac{d\left[ g\left( x \right) \right]}{dx}+g\left( x \right)\dfrac{d\left[ f\left( x \right) \right]}{dx}$ .

Complete step by step answer:
Let us find the first derivative by using the Chain Rule, here f{g(x)} is $\cos \left( {{x}^{2}} \right)$ and g(x) is ${{x}^{2}}$ .
For simplicity, let us use ${{x}^{2}}=t$ .
$\begin{aligned} & \Rightarrow y'=\dfrac{d\cos t}{dt}\dfrac{dt}{dx} \\\ & \Rightarrow y'=\dfrac{d\cos t}{dt}\dfrac{d{{x}^{2}}}{dx} \\\ \end{aligned}$
Thus, the first derivative will be,
$\Rightarrow y'=-\sin t\left( 2x \right)$
as, the derivative of $\cos t$ will be $\sin t$ and the derivative of ${{x}^{2}}$ will be 2x by Power Rule.
Now, replace it with ${{x}^{2}}$ wherever required. Thus we get,
$\Rightarrow y'=-\left( 2x \right)\sin {{x}^{2}}$ .
Now, to get the second derivative of the equation given in the question we again need to differentiate the above expression of y’. For this, we will first apply Product Rule and then the Chain Rule. It will be as follows-
$\Rightarrow \dfrac{d\left[ -\left( 2x \right)\sin {{x}^{2}} \right]}{dx}=-2\dfrac{d\left[ \left( x \right)\sin {{x}^{2}} \right]}{dx}$
as -2 is constant, we took it out of the derivative, now we will use the Product Rule $\dfrac{d\left[ f\left( x \right)g\left( x \right) \right]}{dx}=f\left( x \right)\dfrac{d\left[ g\left( x \right) \right]}{dx}+g\left( x \right)\dfrac{d\left[ f\left( x \right) \right]}{dx}$ in which $f\left( x \right)=x$ and $g\left( x \right)=\sin {{x}^{2}}$ for our question.
So, we will write as,
$\Rightarrow y''=-2\left[ x\dfrac{d\left( \sin {{x}^{2}} \right)}{dx}+\sin {{x}^{2}}\dfrac{dx}{dx} \right]$
which will give us,
$\Rightarrow y''=2\left[ x\cos {{x}^{2}}\left( 2x \right)+\sin {{x}^{2}}\times 1 \right]$
thus, finally giving us,
$\therefore y''=4{{x}^{2}}\cos {{x}^{2}}+2\sin {{x}^{2}}$
Therefore, the second derivative of the equation given in the question is $y''=4{{x}^{2}}\cos {{x}^{2}}+2\sin {{x}^{2}}$ .

Note:
Remember the derivatives of cosx, sinx and ${{x}^{2}}$ . They are –sinx, cosx and 2x respectively. It is important to follow the order to first use the Product Rule and then the Chain Rule to get the second derivative from the first.