Week 4 - The Beginning of Derivatives #

Derivatives #

Definition #

• The derivative of f at point x is defined to be $\displaystyle\lim_{h -> 0}\frac{f(x + h) - f(x)}{h}$

• If the derivative of f exists at x, we say that the function is differentiable at x.

• The derivative is measuring how changing the input affects the output. In other words, how wiggling the input to f changes f.

• Other notations $f’(x) = \frac{d}{dx}f(x) = D_{x}f(x) = \frac{dy}{dx}$.

  • $f’(x)=\displaystyle\lim_{h -> 0}\frac{f(x + h) - f(x)}{h}$

• Different definitions:

  • $f’(x)=\displaystyle\lim_{w -> x}\frac{f(w) - f(x)}{w - x}$
  • $f’(a)=\displaystyle\lim_{x -> a}\frac{f(x) - f(a)}{x - a}$
    • Notice the variable is a not x, cause we are approaching a not x.

• If the derivative of f exists at x whenever x between a and b, then we say that f is differentiable on (a, b).

  • open interval (a, b) can be $(a, \infty)$ or $(-\infty, a)$ or $(-\infty, +\infty)$
  • (a, b) does NOT include a and b.

• Tangent Line

  • the tangent line is the limiting position of the secant line PQ as Q approaches P.
    • 
  • The tangent line to the curve y = f(x) at the point P(a, f(a)) is the line through P with slope $t=\displaystyle\lim_{x -> a}\frac{f(x) - f(a)}{x - a}$ provided that this limit exists.
  • Example: Find an equation of the tangent line to the parabola $y=x^2$ at the point P(1, 1).
    • 

How Can A Function Fail To Be Differentiable (Three situations)? #

• First: the left and right limits are different.
  • Why is f(x) = |x| not differentiable at x = 0 ?
    • $f(x) = |x|$
    • $f’(0) = \displaystyle\lim_{h->0}\frac{|0+h| - |0|}{h}$
    • $f’(0) = \displaystyle\lim_{h->0}\frac{|h|}{h}$ <- DNE(does not exist)
      • Because $\displaystyle\lim_{h->0^{+}}\frac{|h|}{h} = 1 \ne -1 = \displaystyle\lim_{h->0^{-}}\frac{|h|}{h}$
• Second: if f is not continuous at a, then f is not differentiable at a.
• Third: the curve has a vertical tangent line when x = a; that is, f is continuous at a and $\displaystyle\lim_{x->a}|f(x)|=\infty$
  • e.g. $f(x) = \frac{1}{x}$, let x -> 0. check week 2-3’s note to find out what the function look like.

Why Would I Care To Find The Derivative? #

• Why is sqrt(9999) so close to 99.995?
• Let’s assume(will prove later):
  • $\frac{d}{dx}x^{n}=nx^{n-1}$
• Then we can get
  • $\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-\frac{1}{2}}$, same as: $\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$
• we know $\sqrt{10000} = 100$, and $\sqrt{9999} = \sqrt{10000 - 1} \approx 100 - 1 \times (\text{derivative at 10000)}$
• so what is the derivative at 10000? use the derivative function we got above.
  • $f’(10000) = \frac{1}{2\sqrt{10000}} = \frac{1}{2 \times 100} = 0.005$
• then we got $\sqrt{9999} \approx 100 - 1 * 0.005 = 99.995$
• in other words, $\sqrt{9999} \approx 100 - 1 * (\text{derivative at}\\ 10000) = 99.995$
• Conclusion:
  • $f(x+h) \approx f(x)+h*f’(x)$

Rules #

• $\frac{d}{dx}x^{n}=nx^{n-1}$
• $\frac{d}{dx}nf(x)=n\frac{d}{dx}f(x)$
• $\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$
• More in next week

Review Questions #

• definition of derivatives
• what does derivatives measure?
• what is the notion for derivatives?
• How Can A Function Fail To Be Differentiable?
• Calculate $\sqrt{9999}$