Question

Find the second and third derivative of function \[y={{x}^{2}}\sin 2x\].

Answer 1

Hint: Use multiplication rule of differentiation \[\left( \dfrac{d}{dx}\left( u.v \right) \right)=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}\]and take care of chain rule wherever necessary.

Complete step-by-step solution -
Here, we have expression as
\[y={{x}^{2}}\sin 2x-(1)\]
Now, differentiating the given functions or equation (1) as:
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}}\sin 2x \right)$
As we can observe, the given function is multiplication of two functions \[{{x}^{2}}\]and \[\sin 2x\]which are algebraic and trigonometric respectively. Hence, we need to apply multiplication rule as stated below:
If we have two functions $u(x)$ and $v(x)$ in multiplication as $y=u(x).v(x)$, then $\dfrac{dy}{dx}$ can be defined as,
\[\dfrac{dy}{dx}=u(x)\dfrac{dv(x)}{dx}+v(x)\dfrac{du(x)}{dx}-(2)\]
Now, coming to question,
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}}\sin 2x \right)$
We can compare it with equation (2) and apply the multiplication rule as stated, with $u(x)={{x}^{2}}$ and $v(x)=\sin 2x$ .
Hence, we can write $\dfrac{dy}{dx}$ as,
\[\dfrac{dy}{dx}=\dfrac{d}{dx}({{x}^{2}}\sin 2x)={{x}^{2}}\dfrac{d}{dx}(\sin 2x)+\sin 2x\dfrac{d}{dx}({{x}^{2}})\]
Here, we need to take care of chain rule in implicit function as well.
\[\begin{align}
  & h(x)=f(g(x)) \\
 & {{h}^{'}}(x)={{f}^{'}}(g(x)){{g}^{'}}(x) \\
\end{align}\]
It means differentiating function one by one continuously is termed as chain rule of differentiation.
Therefore, applying chain rule in \[\dfrac{dy}{dx}-\]
\[\begin{align}
  & \dfrac{dy}{dx}={{x}^{2}}\cos 2x\dfrac{d}{dx}(2x)+\sin 2x\times 2x \\
 & \dfrac{dy}{dx}=2{{x}^{2}}\cos 2x+2x\sin 2x-(3)\left( \because \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},\dfrac{d}{dx}\sin x=\cos x \right) \\
\end{align}\]
As we need to calculate second derivative and third derivative of the given function, so differentiating equation (3) again for second derivative of given function or $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ :
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}(2{{x}^{2}}\cos 2x)+2\dfrac{d}{dx}(x\sin 2x)\left( \because \dfrac{d}{dx}(f(x)+g(x))={{f}^{'}}(x)+{{g}^{'}}(x) \right) \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\dfrac{d}{dx}({{x}^{2}}\cos 2x)+2\dfrac{d}{dx}(x\sin 2x)-(4) \\
\end{align}\]
Here we can see that we have to calculate $\dfrac{d}{dx}\left( {{x}^{2}}\cos 2x \right)$ and $\dfrac{d}{dx}\left( x\sin x \right)$.
Let we have two functions,
$\begin{align}
  & {{F}_{1}}(x)={{x}^{2}}\cos 2x \\
 & {{F}_{2}}(x)=x\sin 2x \\
\end{align}$
Now, we need to find $\dfrac{d{{F}_{1}}}{dx}$ or $\dfrac{d}{dx}\left( {{x}^{2}}\cos 2x \right)$, here we can observe that ${{x}^{2}}$ and $\cos 2x$ are two functions in multiplication. So, we need to apply the multiplication rule of differentiation as stated in equation (2) with $u={{x}^{2}},v=\cos 2x$ .
Hence,
\[\begin{align}
  & \dfrac{d{{F}_{1}}}{dx}=\dfrac{d}{dx}\left( {{x}^{2}}\cos 2x \right) \\
 & \dfrac{d{{F}_{1}}}{dx}=\cos 2x\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{x}^{2}}\dfrac{d}{dx}\left( \cos 2x \right) \\
 & \dfrac{d{{F}_{1}}}{dx}=\cos 2x(2x)-\sin 2x({{x}^{2}})\dfrac{d}{dx}\left( 2x \right)\left( \because \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}},\dfrac{d}{dx}\left( \cos x \right)=-\sin x \right) \\
\end{align}\]
\[\dfrac{d{{F}_{1}}}{dx}=2x\cos 2x-2{{x}^{2}}\sin 2x...........(5)\]
Similarly, ${{F}_{2}}$ has $x$ and $\sin 2x$ functions in multiplications, hence we need to apply some multiplication rule of differentiation as stated in equation (2).
Here we have $u=x,v=\sin 2x$
\[\begin{align}
  & \dfrac{d{{F}_{2}}}{dx}=\dfrac{d}{dx}\left( x\sin 2x \right) \\
 & \dfrac{d{{F}_{2}}}{dx}=\sin 2x\dfrac{d}{dx}\left( x \right)+x\dfrac{d}{dx}\left( sin2x \right) \\
 & \dfrac{d{{F}_{2}}}{dx}=\sin 2x(1)+x\cos 2x\dfrac{d}{dx}\left( 2x \right)\left( \because \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}},\dfrac{d}{dx}\left( sinx \right)=\cos x \right) \\
\end{align}\]
\[\dfrac{d{{F}_{2}}}{dx}=2x\cos 2x+\sin 2x...........(6)\]
Now putting values of equation (5) and (6) in equation (4), we get
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( 2x\cos x-2{{x}^{2}}\sin 2x \right)+2\left( 2x\cos 2x+\sin 2x \right) \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4x\cos 2x-4{{x}^{2}}\sin 2x+2\sin 2x+4x\cos 2x \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-4{{x}^{2}}\sin 2x+8x\cos 2x+2\sin 2x-(7) \\
\end{align}\]
Now we need to calculate third derivative of the given function ${{x}^{2}}\sin 2x$ ; so now, we have to differentiate equation (7) again:
\[\begin{align}
  & \dfrac{d}{dx}(\dfrac{{{d}^{2}}y}{d{{x}^{2}}})=\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}(-4{{x}^{2}}\sin 2x+8x\cos 2x+2\sin 2x) \\
 & \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-4\dfrac{d}{dx}({{x}^{2}}\sin 2x)+8\dfrac{d}{dx}(x\cos 2x)+2\dfrac{d}{dx}(\sin 2x).......(8) \\
\end{align}\]
Here we have three functions ${{x}^{2}}\sin 2x,x\cos 2x$ and $\cos 2x$ ofwhich we need to find derivative. Let us suppose,
$\begin{align}
  & {{A}_{1}}(x)={{x}^{2}}\sin 2x \\
 & {{A}_{2}}(x)=x\cos 2x \\
 & {{A}_{3}}(x)=\sin 2x \\
\end{align}$
For $\dfrac{d{{A}_{1}}(x)}{dx}$, we can observe the given function in question $y={{x}^{2}}\sin 2x$ which is same as ${{A}_{1}}(x)$. Hence, we can directly put value of it from equation (3)
$\dfrac{d{{A}_{1}}(x)}{dx}=2{{x}^{2}}\cos 2x+2x\sin 2x...........(9)$
For $\dfrac{d{{A}_{2}}(x)}{dx}$, we have two functions $x$ and $\cos 2x$ in multiplication, so we need to apply multiplication rule of differentiation as stated in equation (2). Here, we can assume, $u=x,v=\cos 2x$ .
Hence,
$\begin{align}
  & \dfrac{d}{dx}(x\cos 2x)=\cos 2x\dfrac{d}{dx}(x)+x\dfrac{d}{dx}(\cos 2x) \\
 & \dfrac{d{{A}_{2}}(x)}{dx}=\cos 2x-x\sin 2x\dfrac{d}{dx}(2x) \\
 & \dfrac{d{{A}_{2}}(x)}{dx}=\cos 2x-2x\sin 2x......(10) \\
\end{align}$
For $\dfrac{d{{A}_{3}}(x)}{dx}$, we can directly calculate it in following manner:
$\begin{align}
  & \dfrac{d{{A}_{3}}(x)}{dx}=\dfrac{d}{dx}\left( \sin 2x \right)=\cos 2x\dfrac{d}{dx}(2x) \\
 & \dfrac{d{{A}_{3}}(x)}{dx}=2\cos 2x......(11) \\
\end{align}$
Now putting the value of equation (9), (10) and (11) in equation (8), we get
\[\begin{align}
  & \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-4\left( 2{{x}^{2}}\cos 2x+2x\sin 2x \right)+8\left( \cos 2x-2x\sin 2x \right)+2(2cos2x) \\
 & \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-8{{x}^{2}}\cos 2x-8x\sin 2x+8\cos 2x-16x\sin 2x+4\cos 2x \\
 & \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-8{{x}^{2}}\cos 2x-24x\sin 2x+12\cos 2x \\
\end{align}\]
This is the required answer.

Note: Here one can go wrong while writing $\dfrac{d}{dx}(\sin 2x)$ or $\dfrac{d}{dx}(\cos 2x)$in following way,
$\dfrac{d}{dx}(\sin 2x)=cos2x$ or $\dfrac{d}{dx}(\cos 2x)=-sin2x$
These are wrong as we need to apply chain rule while differentiating above terms.
Chain rule states that if we have two or more combined functions as $f(g(x)$then derivative of $f(g(x))$ can be defined as;
$(f(g(x)))'=f'(g(x)).g'(x)$
Calculation is an important part of this question.