Question

1. find the derivative.

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