The problem asks us to find the derivatives of the following two functions: a) $y = x\sin(2x+1)$ b) $y = (x^2) \cdot 5^x \cdot \cos(\sin^2(x))$

1. Problem Description

The problem asks us to find the derivatives of the following two functions:

a) $y = x\sin(2x+1)$

b) $y = (x^2) \cdot 5^x \cdot \cos(\sin^2(x))$

2. Solution Steps

a) $y = x\sin(2x+1)$

We need to use the product rule and the chain rule to find the derivative of this function. The product rule states that $(uv)' = u'v + uv'$. Here, let $u = x$ and $v = \sin(2x+1)$. Then $u' = 1$. To find $v'$, we use the chain rule. The chain rule states that $(f(g(x)))' = f'(g(x)) \cdot g'(x)$. The derivative of $\sin(x)$ is $\cos(x)$, and the derivative of $2x+1$ is $2$. Therefore, $v' = \cos(2x+1) \cdot 2 = 2\cos(2x+1)$.

Using the product rule:

$y' = (1)\sin(2x+1) + x(2\cos(2x+1))$

$y' = \sin(2x+1) + 2x\cos(2x+1)$

b) $y = (x^2) \cdot 5^x \cdot \cos(\sin^2(x))$

Let $u = x^2$, $v = 5^x$ and $w = \cos(\sin^2(x))$.

We will use the product rule to find the derivative.

$(uvw)' = u'vw + uv'w + uvw'$

$u' = 2x$.

$v' = 5^x \ln(5)$.

$w' = -\sin(\sin^2(x)) \cdot (2\sin(x) \cos(x))$. This uses the chain rule repeatedly.

Then

$y' = (2x) 5^x \cos(\sin^2(x)) + x^2 (5^x \ln(5)) \cos(\sin^2(x)) + x^2 5^x (-\sin(\sin^2(x)) (2\sin(x)\cos(x)))$

$y' = 2x \cdot 5^x \cos(\sin^2(x)) + x^2 \ln(5) 5^x \cos(\sin^2(x)) - 2x^2 5^x \sin(\sin^2(x)) \sin(x) \cos(x)$

3. Final Answer

a) $y' = \sin(2x+1) + 2x\cos(2x+1)$

b) $y' = 2x \cdot 5^x \cos(\sin^2(x)) + x^2 \ln(5) 5^x \cos(\sin^2(x)) - 2x^2 5^x \sin(\sin^2(x)) \sin(x) \cos(x)$