Question

homework4: problem 30
(1 point)
differentiate ( y = 5 sin ( \tan sqrt { sin x } ) ).
( y ^ { prime } = \frac { 5 cos ( \tan ( sqrt { sin ( x ) } sec ^ { 2 } ( sqrt { sin ( x ) } ^ { 2 } ) cos ( x ) ) ) ) } { 2 sqrt { sin ( x ) } } )
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Explanation:

Step1: Apply the chain rule for \(y = 5\sin(u)\) where \(u=\tan(v)\) and \(v = \sqrt{\sin x}\)

The chain rule states that if \(y = f(g(h(x)))\), then \(y^\prime=f^\prime(g(h(x)))\cdot g^\prime(h(x))\cdot h^\prime(x)\). For \(y = 5\sin(u)\), \(y^\prime=5\cos(u)\cdot u^\prime\).

Step2: Differentiate \(u = \tan(v)\)

Using the formula \(\frac{d}{dx}\tan(x)=\sec^{2}(x)\), we get \(u^\prime=\sec^{2}(v)\cdot v^\prime\).

Step3: Differentiate \(v=\sqrt{\sin x}=(\sin x)^{\frac{1}{2}}\)

Using the power - rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\) and the chain rule, \(v^\prime=\frac{1}{2}(\sin x)^{-\frac{1}{2}}\cdot\cos x=\frac{\cos x}{2\sqrt{\sin x}}\)

Step4: Substitute back \(u\) and \(v\)

Substitute \(u = \tan(\sqrt{\sin x})\) and \(v=\sqrt{\sin x}\) into the derivative.
\(y^\prime=5\cos(\tan(\sqrt{\sin x}))\cdot\sec^{2}(\sqrt{\sin x})\cdot\frac{\cos x}{2\sqrt{\sin x}}\)

Answer:

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