Find the derivative of $y=x^{2}$.
$\frac{dy}{dx}=2x$
$\frac{dy}{dx}=x$
y=2x
$\frac{dx}{dy}=2x$
Find the first derivative of the following function:
$f(x)=4x^4-\frac{2}{3}x^3+\frac{3}{2}x^2-17x+43$
$f'(x)=8x^3-4x^2+6x-\frac{17}{2}$
$f'(x)=16x^3-2x^2+3x-17$
$f'(x)=\frac{16}{3}x^3-x^2+\frac{3}{2}x-17$
$f'(x)=4x^3-\frac{2}{3}x^2+\frac{3}{2}x-17$
$f'(x)=x^3+2x^2-3x+34$
