3.6 HW Problems

Table of Contents:

Chain rule (video) | Khan Academy

• How do we differentiate $F(x)=\sin(x^2-4)$?
• This function is the composition $f\circ g$ of two functions
  • (1) $y=f(u)=\sin(u)$
  • and (2) $u=g(x)=x^2-4$
• The answer, given via the Chain Rule, says that the derivative is the product of the derivatives of f and g.

Derivative of a Composite Function

• The function $y=\frac{3}{2}x=\frac{1}{2}(3x)$ is the composition of the functions $y=\frac{1}{2}u$ and $u=3x$. So we have:

$$ \frac{dy}{dx}=\frac{3}{2}, \enspace \frac{dy}{du}=\frac{1}{2}, \enspace \frac{du}{dx}=3 $$

• Since $\frac{3}{2}=\frac{1}{2}\cdot3$, we see in this case that

$$ \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx} $$

• The derivative of the composite function $f(g(x))$ at x is the “derivative of f at g(x) times the derivative of g at x”
  • this is known as the Chain Rule

The Chain Rule

• If f(u) is differentiable at the point $u=g(x)$ and g(x) is differentiable at x, then the composite function $(f \circ g)(x)=f(g(x))$ is differentiable at x, and

$$ (f \circ g)'(x)=f'(g(x)) \cdot g'(x) $$

• Remember the “outside then inside” trick.
  • the HW problems show AND explain the chain rule best!