Differentiate the following functions using the product rule and chain rule: y = x2 x, y = e^x 3x, y = e3x nx, $y = (x

Let $u = x^2$ and $v = \cos x$.

$u' = \frac{d}{dx}(x^2) = 2x$.
$v' = \frac{d}{dx}(\cos x) = -\sin x$.

2) $y = e^x \sin 3x$

Let $u = e^x$ and $v = \sin 3x$.

$u' = \frac{d}{dx}(e^x) = e^x$.
For $v'$, use the chain rule: $\frac{d}{dx}(\sin 3x) = \cos(3x) \cdot \frac{d}{dx}(3x) = 3\cos 3x$.

3) $y = e^{3x} \cos nx$

Let $u = e^{3x}$ and $v = \cos nx$.

For $u'$, use the chain rule: $\frac{d}{dx}(e^{3x}) = e^{3x} \cdot \frac{d}{dx}(3x) = 3e^{3x}$.
For $v'$, use the chain rule: $\frac{d}{dx}(\cos nx) = -\sin(nx) \cdot \frac{d}{dx}(nx) = -n\sin nx$.

4) $y = (x \sin x)^2$

Let $w = x \sin x$. Then $y = w^2$.
$\frac{dy}{dw} = 2w = 2x \sin x$.

Let $u = x$ and $v = \sin x$.
$u' = 1$ and $v' = \cos x$.
$\frac{dw}{dx} = (1)(\sin x) + (x)(\cos x) = \sin x + x \cos x$.

Using the identity $\sin(2x) = 2\sin x \cos x$:

5) $y = e^{3x} \cos 4x$

Let $u = e^{3x}$ and $v = \cos 4x$.

For $u'$, use the chain rule: $\frac{d}{dx}(e^{3x}) = e^{3x} \cdot \frac{d}{dx}(3x) = 3e^{3x}$.
For $v'$, use the chain rule: $\frac{d}{dx}(\cos 4x) = -\sin(4x) \cdot \frac{d}{dx}(4x) = -4\sin 4x$.

6) $y = x^3 \ln(x^2+2)$

Let $u = x^3$ and $v = \ln(x^2+2)$.

$u' = \frac{d}{dx}(x^3) = 3x^2$.
For $v'$, use the chain rule: $\frac{d}{dx}(\ln(x^2+2))$. Let $w = x^2+2$, so $\frac{dw}{dx} = 2x$.
Then $\frac{d}{dx}(\ln w) = \frac{1}{w} \cdot \frac{dw}{dx} = \frac{1}{x^2+2} \cdot 2x = \frac{2x}{x^2+2}$.

You can factor out $x^2$: