# 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 $\displaystyle\lim_{x->a}|f'(x)|=\infty$

### Why Would I Care To Find The Derivative?

• Why is sqrt(9999) so close to 99.995?
• $\frac{d}{dx}x^{n}=nx^{n-1}$
• $\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-\frac{1}{2}}$
• $\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$
• we know that the tangent line of $f(x)=\sqrt{x}$ at 10000 is 1/200 which is 0.005.
• so $0.005*(10000-9999) \approx 100 - \sqrt{9999}$
• then we got $\sqrt{9999} \approx 100 - 0.005 * 1 = 99.995$
• 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