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.

& h(x)=f(g(x)) \\\ & {{h}^{'}}(x)={{f}^{'}}(g(x)){{g}^{'}}(x) \\\ \end{aligned}$$ It means differentiating function one by one continuously is termed as chain rule of differentiation. Therefore, applying chain rule in $$\dfrac{dy}{dx}-$$ $$\begin{aligned} & \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{aligned}$$ As we need to calculate second derivative and third derivative of the given function, so differentiating equation (3) again for second derivative of given function or $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ : $$\begin{aligned} & \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{aligned}$$ Here we can see that we have to calculate $\dfrac{d}{dx}\left( {{x}^{2}}\cos 2x \right)$ and $\dfrac{d}{dx}\left( x\sin x \right)$. Let we have two functions, $\begin{aligned} & {{F}_{1}}(x)={{x}^{2}}\cos 2x \\\ & {{F}_{2}}(x)=x\sin 2x \\\ \end{aligned}$ Now, we need to find $\dfrac{d{{F}_{1}}}{dx}$ or $\dfrac{d}{dx}\left( {{x}^{2}}\cos 2x \right)$, here we can observe that ${{x}^{2}}$ and $\cos 2x$ are two functions in multiplication. So, we need to apply the multiplication rule of differentiation as stated in equation (2) with $u={{x}^{2}},v=\cos 2x$ . Hence, $$\begin{aligned} & \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{aligned}$$ $$\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{aligned} & \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{aligned}$$ $$\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{aligned} & \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{aligned}$$ Now we need to calculate third derivative of the given function ${{x}^{2}}\sin 2x$ ; so now, we have to differentiate equation (7) again: $$\begin{aligned} & \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{aligned}$$ Here we have three functions ${{x}^{2}}\sin 2x,x\cos 2x$ and $\cos 2x$ ofwhich we need to find derivative. Let us suppose, $\begin{aligned} & {{A}_{1}}(x)={{x}^{2}}\sin 2x \\\ & {{A}_{2}}(x)=x\cos 2x \\\ & {{A}_{3}}(x)=\sin 2x \\\ \end{aligned}$ For $\dfrac{d{{A}_{1}}(x)}{dx}$, we can observe the given function in question $y={{x}^{2}}\sin 2x$ which is same as ${{A}_{1}}(x)$. Hence, we can directly put value of it from equation (3) $\dfrac{d{{A}_{1}}(x)}{dx}=2{{x}^{2}}\cos 2x+2x\sin 2x...........(9)$ For $\dfrac{d{{A}_{2}}(x)}{dx}$, we have two functions $x$ and $\cos 2x$ in multiplication, so we need to apply multiplication rule of differentiation as stated in equation (2). Here, we can assume, $u=x,v=\cos 2x$ . Hence, $\begin{aligned} & \dfrac{d}{dx}(x\cos 2x)=\cos 2x\dfrac{d}{dx}(x)+x\dfrac{d}{dx}(\cos 2x) \\\ & \dfrac{d{{A}_{2}}(x)}{dx}=\cos 2x-x\sin 2x\dfrac{d}{dx}(2x) \\\ & \dfrac{d{{A}_{2}}(x)}{dx}=\cos 2x-2x\sin 2x......(10) \\\ \end{aligned}$ For $\dfrac{d{{A}_{3}}(x)}{dx}$, we can directly calculate it in following manner: $\begin{aligned} & \dfrac{d{{A}_{3}}(x)}{dx}=\dfrac{d}{dx}\left( \sin 2x \right)=\cos 2x\dfrac{d}{dx}(2x) \\\ & \dfrac{d{{A}_{3}}(x)}{dx}=2\cos 2x......(11) \\\ \end{aligned}$ Now putting the value of equation (9), (10) and (11) in equation (8), we get $$\begin{aligned} & \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{aligned}$$ 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.