Question

Find the derivative of $\dfrac{{{d}^{2}}\left( 2\cos x\cos 3x \right)}{d{{x}^{2}}}=$
$1){{2}^{2}}\left( {{2}^{2}}\cos 4x+\cos 2x \right)$
$2){{2}^{2}}\left( -{{2}^{2}}\cos 4x+\cos 2x \right)$
$3){{2}^{2}}\left( {{2}^{2}}\cos 4x-\cos 2x \right)$
$4)-{{2}^{2}}\left( {{2}^{2}}\cos 4x+\cos 2x \right)$

Answer 1

Hint: To solve this question we need to have the knowledge of differentiation. In the question given above the function is in terms of the product of two trigonometric functions $\cos x$ and $\cos 3x$. So to solve the problem we will use the formula of differentiation for $u$ and $v$ which implies $\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ . On solving this we get the first differentiation while we will differentiate again to find the second order differentiation of the result we got.

Complete step-by-step solution:
The question asks us to find the double differentiation of the function given to us which is $2\cos x\cos 3x$. Now we will solve the problem by considering this function to be the product of the two other functions $u$ and $v$ where $u=\cos x$ and $v=\cos 3x$. To find the differentiation we will use the formula of the product of the two function which is $\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. On applying the same we get:
$\Rightarrow \dfrac{d\left( 2\cos x\cos 3x \right)}{dx}=2\cos x\dfrac{d\cos 3x}{dx}+\cos 3x\dfrac{d2\cos x}{dx}$
$\Rightarrow \dfrac{d\left( 2\cos x\cos 3x \right)}{dx}=2\cos x\dfrac{d\cos 3x}{dx}+2\cos 3x\dfrac{d\cos x}{dx}$
As per the formula of differentiation we know that differentiation of the trigonometric function $\cos x$ is $-\sin x$ while differentiation of $\cos 3x$ is $-3\sin 3x$. On applying the same in the above expression we get:
$\Rightarrow \dfrac{d\left( 2\cos x\cos 3x \right)}{dx}=2\cos x\left( -3\sin 3x \right)+2\cos 3x\left( -\sin x \right)$
$\Rightarrow \dfrac{d\left( 2\cos x\cos 3x \right)}{dx}=-6\cos x\sin 3x-2\cos 3x\sin x$
$\Rightarrow \dfrac{d\left( 2\cos x\cos 3x \right)}{dx}=-3\left[ 2\cos x\sin 3x \right]-1\left[ 2\cos 3x\sin x \right]$
Now we will apply the formula of $2\sin a\cos b=\sin \left( a+b \right)+\sin \left( a-b \right)$ in the above expression. On doing this we get:
$\Rightarrow \dfrac{d\left( 2\cos x\cos 3x \right)}{dx}=-3\left[ \sin 4x+\sin 2x \right]-1\left[ \sin 4x+\sin (-2x) \right]$
$\Rightarrow \dfrac{d\left( 2\cos x\cos 3x \right)}{dx}=-3\sin 4x-3\sin 2x-\sin 4x+\sin 2x$
$\Rightarrow \dfrac{d\left( 2\cos x\cos 3x \right)}{dx}=-4\sin 4x-2\sin 2x$
$\Rightarrow \dfrac{dy}{dx}=-4\sin 4x-2\sin 2x$
The above is the result of one order differentiation, we will again differentiate it to find the double differentiation of the function given in the question. On again differentiating it we get:
$\Rightarrow \dfrac{d\left( \dfrac{dy}{dx} \right)}{dx}=\dfrac{d\left( -4\sin 4x-2\sin 2x \right)}{dx}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( -4\sin 4x \right)+d\left( -2\sin 2x \right)}{dx}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{d\left( 4\sin 4x \right)}{dx}-\dfrac{d\left( 2\sin 2x \right)}{dx}$
As per the formula of differentiation which says $\sin ax$ is $a\cos ax$. On applying the same trigonometric function $\sin x$ is $\cos x$ while differentiation of $\sin 3x$ is $3\cos 3x$. On applying the same in the above expression we get:
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-16\cos 4x-4\cos 2x$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{2}^{2}}\left[ {{2}^{2}}\cos 4x+\cos 2x \right]$
$\therefore $ $\dfrac{{{d}^{2}}\left( 2\cos x\cos 3x \right)}{d{{x}^{2}}}$ is $4)-{{2}^{2}}\left( {{2}^{2}}\cos 4x+\cos 2x \right)$.

Note: To solve this type of question we need to know the formula for the differentiation of the trigonometric function. To find the double differentiation we will have to differentiate the result of the first order differentiation.