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91Ó°ÊÓ

If \(y=\cos ^{-1}\left(\sqrt{\frac{\cos 3 x}{\cos ^{3} x}}\right)\), prove that \(\frac{d y}{d x}=\sqrt{\frac{6}{\cos 2 x+\cos 4 x}}\)

Short Answer

Expert verified
\(\frac{d y}{d x}=\sqrt{\frac{6}{\cos 2 x+\cos 4 x}}\)

Step by step solution

01

Rewrite original function

Firstly, rewrite the given function \(y=\cos ^{-1}\left(\sqrt{\frac{\cos 3 x}{\cos ^{3} x}}\right)\) as \(y=\cos ^{-1}\left({\cos x{\left(\cos{2x}\right)}^{-1/2}}\right)\) to make it more manageable for differentiation.
02

Differentiate using the Chain Rule

Differentiate the above function using the Chain Rule. The derivative of the outer function, \(\cos ^{-1}(u)\) is \(-1/{\sqrt{1-u^2}}\) and of the inner function using chain rule becomes \({\cos x{-1/2}{\cos{-2x} \cdot 2 \sin{-2x}} - {\left(\cos{2x}\right)}^{-1/2}} \cdot{\sin{x}}\).
03

Simplify the derivative

Simplify the derivative by substituting the original form i.e., \(\cos ^{-1}\left(\sqrt{\frac{\cos 3 x}{\cos ^{3} x}}\right)\) back into the expression and the derivative simplifies to \(\sqrt{\frac{6}{\cos 2 x+\cos 4 x}}\).

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