Question

evaluate the derivative of the following function.
f(w)=\cos \sin ^{-1}(8w)
f(w)=\square

Explanation:

Step1: Apply the chain rule

The chain rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \(u = \sin^{- 1}(8w)\), so \(f(w)=\cos(u)\). The derivative of \(\cos(u)\) with respect to \(u\) is \(-\sin(u)\), and the derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1 - x^{2}}}\). Here \(x = 8w\), so \(g(w)=\sin^{-1}(8w)\) and \(g^\prime(w)=\frac{8}{\sqrt{1-(8w)^{2}}}\)

Step2: Substitute back

Since \(u = \sin^{-1}(8w)\), \(f^\prime(w)=-\sin(\sin^{-1}(8w))\cdot\frac{8}{\sqrt{1 - 64w^{2}}}\). But \(\sin(\sin^{-1}(a))=a\) for \(a\in[- 1,1]\), so \(\sin(\sin^{-1}(8w)) = 8w\)

Answer:

\(f^\prime(w)=-\frac{64w}{\sqrt{1 - 64w^{2}}}\)