5. find the derivative.\na. ( y = cos(x^{2}e^{x}) )\nb. ( y = sqrt{5x+\tan(4x)} )

Explanation:

Step1: Differentiate (y = \cos(x^{2}e^{x})) using the chain rule

The chain rule states that if (y=\cos(u)) and (u = x^{2}e^{x}), then (y^\prime=-\sin(u)\cdot u^\prime). First, find (u^\prime) for (u = x^{2}e^{x}) using the product rule. The product rule: ((fg)^\prime=f^\prime g+fg^\prime), where (f = x^{2}), (f^\prime=2x) and (g = e^{x}), (g^\prime=e^{x}). So (u^\prime=(x^{2}e^{x})^\prime=2x e^{x}+x^{2}e^{x}=x e^{x}(2 + x)). Then (y^\prime=-\sin(x^{2}e^{x})\cdot(x e^{x}(2 + x))=-x e^{x}(x + 2)\sin(x^{2}e^{x}))

Step2: Differentiate (y=\sqrt{5x+\tan(4x)}=(5x+\tan(4x))^{\frac{1}{2}}) using the chain rule

Let (u = 5x+\tan(4x)), so (y = u^{\frac{1}{2}}). By the chain rule (y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime). Find (u^\prime): ((5x)^\prime = 5) and ((\tan(4x))^\prime=\sec^{2}(4x)\cdot4) (using the chain rule for (\tan(v)) where (v = 4x)). So (u^\prime=5 + 4\sec^{2}(4x)). Then (y^\prime=\frac{5 + 4\sec^{2}(4x)}{2\sqrt{5x+\tan(4x)}})

Answer:

a. (y^\prime=-x e^{x}(x + 2)\sin(x^{2}e^{x})) b. (y^\prime=\frac{5 + 4\sec^{2}(4x)}{2\sqrt{5x+\tan(4x)}})