# Chapter 14: Derivatives

Recall,

﻿$f'(x) = \lim {h \rightarrow 0}_{f(x+h) - f(x) }{(x+h)-x} = \frac{rise}{run}$﻿

﻿$=$﻿ "The derivative of ﻿$f$﻿ at ﻿$x$﻿"

﻿$=$﻿ "The slope of ﻿$f$﻿ at ﻿$x$﻿"

#### Example

Find the slope of ﻿$f(x) = \frac{1}{2x^2}$﻿ at ﻿$x=3$﻿

Apply the definition of slope

﻿$f'(3) = \lim_{h \rightarrow c} \frac{f(3+h) - f(3)}{(3+h) - 3}$﻿

﻿$= \lim_{h \rightarrow 0} \frac{ f(3+h)-f(3)}{h}$﻿

﻿$= \lim_{h \rightarrow 0} \frac{ \frac{1}{2}(3+h)2- \frac{1}{2}(3)2 }{ h}$﻿

﻿$= \lim_{h \rightarrow 0} \frac{ \frac{1}{2}(3^2+2 \cdot 3 \cdot h+h^2)2- \frac{1}{2}(3)2 }{ h}$﻿

﻿$= \lim_{h \rightarrow 0} \frac{ \frac{1}{2}(6h+h^2) }{ h}$﻿

﻿$= \lim_{h \rightarrow 0} \frac{1}{2} \cdot 6 + \frac{1}{2} h$﻿

﻿$=3$﻿

Thus, the slope at ﻿$x=3$﻿ is ﻿$f'(3)=3$﻿

### Rate of Change

The physical interpretation of derivatives is that they measure rates of change.

#### Definition

If the distance to an object at time ﻿$t$﻿ is ﻿$d(f)$﻿ then its: VELOCITY ﻿$=d'(+) =s(t)$﻿

SPEED ﻿$=|d'(t)| \rightarrow$﻿ Speed ignores direction

ACCELERATION ﻿$= s'(f)$﻿

#### Discuss

An object dropped near Earth travels ﻿$d(t) = 4.9t^2$﻿ meters in ﻿$t$﻿ seconds. How far do you need to drop an object so that it hits the ground at ﻿$9.8m/s$﻿?

Find speed at time ﻿$t$﻿

﻿$d'(t) =9.8t$﻿

Solve for ﻿$d'(t) =9.8$﻿

﻿$d'(t) = 9.8t =9.8 \Rightarrow1$﻿

Find the distance traveled at ﻿$t =1$﻿

﻿$d(1)=4.9t^2 =4.9m$﻿

Thus, we must drop the object from a height of 4.9m.

#### ExamplE

Gregor Mendel discovered the laws of ''genetic inheritance'' by studying populations of peas.

He found: If ﻿$p$﻿ is the percentage of smooth skinned peas in a population then the proportion in the next generation will be:

﻿$s= 2p(1-p)=2p-2p^2$﻿

When does ﻿$s$﻿ change most as a function of ﻿$p$﻿?

Find ﻿$\frac {ds}{dp}$﻿

﻿$\frac{ds}{dp} = \lim_{h \rightarrow 0} \frac{s(p+h) -s(p)}{(p+h)-p}$﻿

﻿$= \lim_{h \rightarrow 0} \frac {s(p+h)-s(p)}{h}$﻿

﻿$= \lim_{h \rightarrow 0} \frac {2(p+h)-2(p+h)^2]-[2p-2p^2]}{h}$﻿

﻿$= \lim_{h \rightarrow 0} \frac {2h-4ph-2h^2}{h}$﻿

﻿$= \lim_{h \rightarrow 0} 2-4p-2h$﻿

﻿$=2-4p$﻿

Thus, the proportion changes most when ﻿$p=0$﻿

That is, introducing a few smooth skin peas will have the largest affect on the population

We need a more systematic way of calculating derivatives.

### Differentiation Rules

#### Fact

If ﻿$f(x) =k$﻿ for all ﻿$x$﻿ then ﻿$f'(x)=0$﻿

#### Discuss

Why is this true?

﻿$f'(x) = \lim_{h \rightarrow 0 } \frac{f(x+h)-f(x)}{h}$﻿ ﻿$\rightarrow$﻿ Substitute ﻿$f(x)=k$﻿

﻿$= \lim_{h \rightarrow 0 } \frac{k-k}{h}$﻿

﻿$\frac{d}{dx}(k) =0$﻿

#### Fact

If ﻿$f(x)= c \cdot g(x)$﻿ for some number ﻿$c$﻿ then ﻿$f'(x) = c \cdot g'(x)$﻿

#### Discuss

Why is this true ? ﻿$\rightarrow$﻿ Transformation

﻿$f'(x) = \lim_{h \rightarrow 0} \frac { f(x+h)-f(x)}{h}$﻿ ﻿$\rightarrow$﻿ Substitute ﻿$f(x) = c \cdot g(x)$﻿

﻿$= \lim_{h \rightarrow 0} \frac {c \cdot g(x+h)-c \cdot g(x)}{h}$﻿

﻿$=c \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h }$﻿ ﻿$\rightarrow$﻿ Limit Law

﻿$=cg'(x)$﻿

﻿$\frac{d}{dx}[c \cdot g(x)] =c \frac {d}{dx}[g(x)]$﻿

### Summary

Constants ﻿$\rightarrow$﻿ ﻿$f(x)= k \Rightarrow f'(x)=0$﻿

Scaling ﻿$\rightarrow$﻿ ﻿$f(x)=c \cdot g(x) \Rightarrow f'(x)= c \cdot g'(x)$﻿