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