Find the equation of the tangent to the curve y = 5 − 2x − x2 at the point (1, 2). Use the quotient rule to differentiate y = 3x/(x2+1).

Let $u = 3x+x^4$. Then $u' = \frac{d}{dx}(3x+x^4) = 3 + 4x^3$.
Let $v = 2x^2+1$. Then $v' = \frac{d}{dx}(2x^2+1) = 4x$.

Numerator:
$(2x^2+1)(3+4x^3) = 6x^2 + 8x^5 + 3 + 4x^3 = 8x^5 + 4x^3 + 6x^2 + 3$
$(3x+x^4)(4x) = 12x^2 + 4x^5$
Subtracting the second from the first:
$(8x^5 + 4x^3 + 6x^2 + 3) - (12x^2 + 4x^5)$
$= 8x^5 + 4x^3 + 6x^2 + 3 - 12x^2 - 4x^5$
$= 4x^5 + 4x^3 - 6x^2 + 3$

Let $u = 3x^2$. Then $\frac{du}{dx} = 6x$.
The derivative of $5 \sin(u)$ with respect to $u$ is $5 \cos(u)$.

Let $u = \sqrt{x} = x^{1/2}$. Then $\frac{du}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
The derivative of $4 \tan(u)$ with respect to $u$ is $4 \sec^2(u)$.