Solve 80,82 and 83 and 85

The product rule states that if $y = uv$, then $\frac{dy}{dx} = u'v + uv'$.
Let $u = x^2$ and $v = \sin x$.
Then $u' = \frac{d}{dx}(x^2) = 2x$ and $v' = \frac{d}{dx}(\sin x) = \cos x$.

$$\frac{dy}{dx} = (2x)(\sin x) + (x^2)(\cos x)$$

$$\frac{dy}{dx} = 2x \sin x + x^2 \cos x$$ $$\boxed{\frac{dy}{dx} = 2x \sin x + x^2 \cos x}$$

82. Find $\frac{dy}{dx}$ if $y = \frac{x^2}{x+1}$.

The quotient rule states that if $y = \frac{f(x)}{g(x)}$, then $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$.
Let $f(x) = x^2$ and $g(x) = x+1$.
Then $f'(x) = \frac{d}{dx}(x^2) = 2x$ and $g'(x) = \frac{d}{dx}(x+1) = 1$.

$$\frac{dy}{dx} = \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2}$$

$$\frac{dy}{dx} = \frac{2x^2 + 2x - x^2}{(x+1)^2}$$ $$\frac{dy}{dx} = \frac{x^2 + 2x}{(x+1)^2}$$ $$\boxed{\frac{dy}{dx} = \frac{x^2 + 2x}{(x+1)^2}}$$

83. Find $\frac{dy}{dx}$ if $y = \sqrt{x^2+1}$.

Rewrite $y = \sqrt{x^2+1}$ as $y = (x^2+1)^{1/2}$.
The chain rule states that if $y = [f(x)]^n$, then $\frac{dy}{dx} = n[f(x)]^{n-1}f'(x)$.
Here, $f(x) = x^2+1$ and $n = \frac{1}{2}$.
So, $f'(x) = \frac{d}{dx}(x^2+1) = 2x$.

$$\frac{dy}{dx} = \frac{1}{2}(x^2+1)^{\frac{1}{2}-1} \cdot (2x)$$ $$\frac{dy}{dx} = \frac{1}{2}(x^2+1)^{-\frac{1}{2}} \cdot (2x)$$

$$\frac{dy}{dx} = x(x^2+1)^{-\frac{1}{2}}$$ $$\frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}}$$ $$\boxed{\frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}}}$$

85. Find $\frac{dy}{dx}$ if $x^2 + y^2 = 25$.

Remember that $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$ by the chain rule.

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$ $$2x + 2y \frac{dy}{dx} = 0$$

Subtract $2x$ from both sides:

$$2y \frac{dy}{dx} = -2x$$

Divide by $2y$:

$$\frac{dy}{dx} = \frac{-2x}{2y}$$

$$\frac{dy}{dx} = -\frac{x}{y}$$ $$\boxed{\frac{dy}{dx} = -\frac{x}{y}}$$