# Chain rule and implicit differentiation

Theorem For Chain Rule
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x then y = f(g(x)) is a differentiable function of x and
$\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}$
or equivalently,
$\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$

Example
Find
$\frac{dy}{dx}$ for
$y = (x^2+1)^3$

In this case, you would divide the equation as follows:
$\frac{dy}{du} = 3(x^2+1)^2$
and
$\frac{du}{dx} = (2x)$
$\frac{dy}{dx} = 3(x^2+1)^2(2x)$
(Larson Hostetler Edwards) Section 2.4, p.131

Guidelines for Implicit Differentiation
Differentiate both sides of the equation with respect to x.
Collect all terms involving dy/dx on the left side of the equation and move all other terms to the right side of the equation.
Factor
$\frac{dy}{dx}$ out of the left side of the equation.
Solve for
$\frac{dy}{dx}$

Example
Find
$\frac{dy}{dx}$
given that
$y^3+y^2-5y-x^2 = -4$
$\frac{d}{dx}[y^3+y^2-5y-x^2] = \frac{d}{dx}[-4]$
$3y^2\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x = 0$
$3y^2\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx} = 2x$
$\frac{dy}{dx}(3y^2+2y-5) = 2x$
$\frac{dy}{dx} = (2x)/(3y^2+2y-5)$
(Larson Hostetler Edwards) Section 2.5, p.141

Need more examples?

Larson, Ron, Robert P. Hostetler, and Bruce H. Edwards. Calculus. 8th ed. Boston and New York: Houghton Mifflin Company, 2006.