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

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): [ \begin{align*} y'&=5\cos(\tan\sqrt{\sin x})\times\sec^{2}(\sqrt{\sin x})\times\frac{\cos x}{2\sqrt{\sin x}}\ &=\frac{5\cos(\tan\sqrt{\sin x})\sec^{2}(\sqrt{\sin x})\cos x}{2\sqrt{\sin x}} \end{align*} ]

Answer:

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