Question

If \[y = cos\left( {3co{s^{ - 1}} x} \right)\]. Then what is the value of \[\dfrac{{{d^3}y}}{{d{x^3}}}\]?
A. \[24\]
B. \[27\]
C. \[3 - 12{x^2}\]
D. \[ - 24\]

Answer 1

Hint: Simplify the given trigonometric equation by using the formula of \[\cos 3x\]. After that, differentiate the equation three times with respect to \[x\] and get the required answer.

Formula used:
1. \[\cos\left( {3x} \right) = 4\cos^{3} x – 3\cos x\]
2. \[\cos\left( {\cos^{ - 1} x} \right) = x\]
3. \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{\left( {n - 1} \right)}}\]

Complete step by step solution:
The given trigonometric equation is \[y = \cos\left( {3cos^{ - 1} x} \right)\].
Let’s simplify the equation in terms of \[x\].
Apply the formula \[\cos\left( {3x} \right) = 4\cos^{ 3} x – 3\cos x\].
\[y = 4{\left( {cos\left( {co{s^{ - 1}} x} \right)} \right)^3} - 3cos\left( {co{s^{ - 1}} x} \right)\]
\[ \Rightarrow \]\[y = 4{\left( x \right)^3} - 3\left( x \right)\] [ Since \[cos\left( {\cos^{ - 1} x} \right) = x\]]
Now differentiate the above equation with respect to \[x\].
Apply the differentiation formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{\left( {n - 1} \right)}}\].
\[\dfrac{{dy}}{{dx}} = 4 \times 3{\left( x \right)^2} - 3\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = 12{x^2} - 3\]
Now differentiate the first derivative with respect to \[x\].
\[\dfrac{{{d^2}y}}{{d{x^2}}} = \left( {12 \times 2} \right)x - 0\]
\[ \Rightarrow \]\[\dfrac{{{d^2}y}}{{d{x^2}}} = 24x\]
Differentiate the second derivative with respect to \[x\].
\[\dfrac{{{d^3}y}}{{d{x^3}}} = 24\]
Hence the correct option is A.

Note: Students often start the solution to find the derivative of \[y = \cos \left( {3{{\cos }^{ - 1}}x} \right)\] by using the chain rule. It is a long and wrong procedure. We will treat \[{\cos ^{ - 1}}x\] as \[\theta \]. Then we will apply \[\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \] and simplify it. Now we can differentiate it.