Week 4 - The Beginning of Derivatives

1. Derivatives

1.1. Definition

  • The derivative of f at point x is defined to be limh>0f(x+h)f(x)h\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)=ddxf(x)=Dxf(x)=dydxf'(x) = \frac{d}{dx}f(x) = D_{x}f(x) = \frac{dy}{dx}.

    • f(x)=limh>0f(x+h)f(x)hf'(x)=\displaystyle\lim_{h -> 0}\frac{f(x + h) - f(x)}{h}
  • Different definitions:

    • f(x)=limw>xf(w)f(x)wxf'(x)=\displaystyle\lim_{w -> x}\frac{f(w) - f(x)}{w - x}
    • f(a)=limx>af(x)f(a)xaf'(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,)(a, \infty) or (,a)(-\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 m=limx>af(x)f(a)xam=\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=x2y=x^2 at the point P(1, 1).

1.2. How Can A Function Fail To Be Differentiable?

  • First: the left and right limits are different.
    • Why is f(x) = |x| not differentiable at x = 0 ?
      • f(x)=xf(x) = |x|
      • f(0)=limh>00+h0hf'(0) = \displaystyle\lim_{h->0}\frac{|0+h| - |0|}{h}
      • f(0)=limh>0hhf'(0) = \displaystyle\lim_{h->0}\frac{|h|}{h} <- DNE(does not exist)
        • Because limh>0+hh=11=limh>0hh\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 limx>af(x)=lim_{x->a}|f'(x)|=\infty

1.3. Why Would I Care To Find The Derivative?

  • Why is sqrt(9999) so close to 99.995?
    • ddxxn=nxn1\frac{d}{dx}x^{n}=nx^{n-1}
    • ddxx1/2=12x12\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-\frac{1}{2}}
    • ddxx=12x\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}
    • we know that the tangent line of f(x)=xf(x)=\sqrt{x} at 10000 is 1/200 which is 0.005.
    • so 0.005(100009999)10099990.005*(10000-9999) \approx 100 - \sqrt{9999}
    • then we got 99991000.0051=99.995\sqrt{9999} \approx 100 - 0.005 * 1 = 99.995
    • in other words, 99991001(derivative at 10000)=99.995\sqrt{9999} \approx 100 - 1 * (derivative\ at\ 10000) = 99.995
    • Conclusion:
      • f(x+h)f(x)+hf(x)f(x+h) \approx f(x)+h*f'(x)

1.4. Rules

  • ddxxn=nxn1\frac{d}{dx}x^{n}=nx^{n-1}
  • ddxnf(x)=nddxf(x)\frac{d}{dx}nf(x)=n\frac{d}{dx}f(x)
  • ddx(f(x)+g(x))=ddxf(x)+ddxg(x)\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)
  • More in next week

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