Question

differentiate ( y = 5sin(\tansqrt{sin x}) ).
( y=\frac{5cos(\tan(sqrt{sin(x)}sec(sqrt{sin(x)}^2))cos(x))}{2sqrt{sin(x)}} )

Explanation:

Step1: Apply the chain rule

Let \(u = \tan\sqrt{\sin x}\), then \(y = 5\sin u\). The derivative of \(y\) with respect to \(u\) is \(y'_u=5\cos u\).

Step2: Differentiate \(u\) with respect to \(x\)

Let \(v = \sqrt{\sin x}\), then \(u=\tan v\). The derivative of \(u\) with respect to \(v\) is \(u'_v=\sec^{2}v\).

Step3: Differentiate \(v\) with respect to \(x\)

\(v = (\sin x)^{\frac{1}{2}}\), using the power - chain rule: \(v'_x=\frac{1}{2}(\sin x)^{-\frac{1}{2}}\cos x=\frac{\cos x}{2\sqrt{\sin x}}\)

Step4: Combine using the chain rule \(y'_x=y'_u\times u'_v\times v'_x\)

Substitute \(u = \tan\sqrt{\sin x}\) and \(v=\sqrt{\sin x}\) into \(y'_x\):

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Answer:

\(y'=\frac{5\cos(\tan\sqrt{\sin x})\sec^{2}(\sqrt{\sin x})\cos x}{2\sqrt{\sin x}}\)