homework4: problem 30\n(1 point)\ndifferentiate ( y = 5 sin ( \tan sqrt { sin x } ) ).\n( y ^ { prime } =…

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}})