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« Product and Quotient Rule for Derivatives
Differentiation: (lesson 3 of 3)

Chain Rule

If y=f(u)y = f(u) is a differentiable function of uu, and u=g(x)u = g(x) is a differentiable function of xx, then y=f(g(x))y = f(g(x)). This a differentiable function of xx, and

dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

or, equivalently,

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)

Example 1:

y=(x2+1)3, dydx=?g(x)=u=x2+1,y=f(u)=u3,dfdx=dfdududx=(3u2)(2x)=3(x2+1)2(2x)=6x(x2+1)2 \begin{aligned} y &= (x^2 + 1)^3, \ \frac{dy}{dx} = ? \\ g(x) &= u = x^2 + 1, \\ y &= f(u) = u^3, \\ \frac{df}{dx} &= \frac{df}{du} \cdot \frac{du}{dx} = (3u^2) (2x) = 3 (x^2 + 1)^2 (2x) = 6x (x^2 + 1)^2 \end{aligned}

The General Power Rule

ddx[u(x)]n=n[u(x)]n1u(x) \frac{d}{dx} [u(x)]^n = n [u(x)]^{n-1} \cdot u' (x)