Question

FInd the derivative of $\dfrac{{{d}^{20}}\left( 2\cos x\cos 3x \right)}{d{{x}^{20}}}=$
a). ${{2}^{20}}\left( \cos 2x-{{2}^{20}}\cos 4x \right)$
b). ${{2}^{20}}\left( \cos 2x+{{2}^{20}}\cos 4x \right)$
c). ${{2}^{20}}\left( \sin 2x+{{2}^{20}}\sin 4x \right)$
d). ${{2}^{20}}\left( \sin 2x-{{2}^{20}}\sin 4x \right)$

Answer 1

To find the 20th derivative of the given trigonometric function, we will simplify it using the trigonometric equation so that we can easily differentiate it. Then there will be a pattern that will be followed while differentiating the function, which we had to recognise and use it to find what will be the 20th derivative.

Complete step-by-step solution:
Moving ahead with the question we had $2\cos x\cos 3x$, whose 20th derivative we need to find out. So let us first simplify it, so that we can easily differentiate it.
So as we know that $2\cos a\cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$. So by comparing it with our question $2\cos x\cos 3x$ we had values of ‘a’ and ‘b’ equal to ‘x’ and ‘3x’. So by using the identity $2\cos a\cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$, we can write $2\cos x\cos 3x$ it as;
$2\cos x\cos 3x=\cos \left( x+3x \right)+\cos \left( x-3x \right)$
Which on further simplifying we will get;
$2\cos x\cos 3x=\cos \left( 4x \right)+\cos \left( -2x \right)$
Since we know that $\cos (-a)=\cos a$ so we can write $\cos (-2x)$ as $\cos 2x$, so we will get;
$2\cos x\cos 3x=\cos \left( 4x \right)+\cos \left( 2x \right)$.
So now we can differentiate it easily; so we have $\cos \left( 4x \right)+\cos \left( 2x \right)$ whose 20th derivative we need to find out its derivative and see if we can find any pattern that is being followed in differentiation;
So let us first find out 1st derivative using the chain rule, which will be;

& \cos \left( 4x \right)+\cos \left( 2x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=-4\sin 4x-2\sin 2x=-2\left( 2\sin 4x+\sin 2x \right) \\\ \end{aligned}$$ Now find the second derivative, so we will get; $\begin{aligned} & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-16\cos 4x-4\cos 2x=-{{2}^{2}}\left( {{2}^{2}}\cos 4x+\cos 2x \right) \\\ & \\\ \end{aligned}$ Till know we did not got any pattern, let us move further to find the derivation till we get the pattern; so let us find out 3rd derivative which will be; $\begin{aligned} & \Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=64\sin 4x+8\sin 2x={{2}^{3}}\left( {{2}^{3}}\sin 4x+\sin 2x \right) \\\ & \\\ \end{aligned}$ Now let us find out 4th derivative, which will be; $\begin{aligned} & \Rightarrow \dfrac{{{d}^{4}}y}{d{{x}^{4}}}=256\cos 4x+16\cos 2x={{2}^{4}}\left( {{2}^{4}}\cos 4x+\cos 2x \right) \\\ \end{aligned}$ As we can see that in the 4th derivative we got the 4th derivative as same as the question with an additional simultaneously addition of the constant term added in the $\cos 4x$ and whole term with ${{2}^{n}}$ in which ‘n’ is the ${{n}^{th}}$ derivative. So we can say that after every 4th term we will get the derivative with an addition of a constant term which is equal to exponential of 2 whose power is equal to ${{n}^{th}}$ derivative. So we can say that after 4th derivative, we will get 8th derivative which will come when we will move with 4 times derivation, so by pattern as which get repeated by every 4 times differentiation , which will be 8th derivative give us same answer, with an addition of constant term, So 8th derivative will be; $\begin{aligned} & \Rightarrow \dfrac{{{d}^{8}}y}{d{{x}^{8}}}={{2}^{8}}\left( {{2}^{8}}\cos 4x+\cos 2x \right) \\\ & \\\ \end{aligned}$ Similarly there will be 12th which will be 4 derivation ahead of 8th derivative, so which will be equal to’ $\begin{aligned} & \Rightarrow \dfrac{{{d}^{12}}y}{d{{x}^{12}}}={{2}^{12}}\left( {{2}^{12}}\cos 4x+\cos 2x \right) \\\ & \\\ \end{aligned}$ After the 12th derivative we will get 16. And after 16 we will get 20, which we need to find out. So 20th derivative of function will be; $\begin{aligned} & \Rightarrow \dfrac{{{d}^{20}}y}{d{{x}^{20}}}={{2}^{20}}\left( {{2}^{20}}\cos 4x+\cos 2x \right) \\\ & \\\ \end{aligned}$ **Hence the answer is ${{2}^{20}}\left( {{2}^{20}}\cos 4x+\cos 2x \right)$, that is option B is correct.** **Note:** For such question in which we need to find the ${{n}^{th}}$ derivative, and value of ‘n’ is large then there will be some pattern which will be followed while differentiation, which we have in our question as every 4th derivation is repeated.