Chapter 14: Derivatives


f(x)=limh0f(x+h)f(x)(x+h)x=riserunf'(x) = \lim {h \rightarrow 0}_{f(x+h) - f(x) }{(x+h)-x} = \frac{rise}{run}

== "The derivative of ff at xx"

== "The slope of ff at xx"


Find the slope of f(x)=12x2 f(x) = \frac{1}{2x^2} at x=3x=3

Apply the definition of slope

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

=limh0f(3+h)f(3)h = \lim_{h \rightarrow 0} \frac{ f(3+h)-f(3)}{h}

=limh012(3+h)212(3)2h = \lim_{h \rightarrow 0} \frac{ \frac{1}{2}(3+h)2- \frac{1}{2}(3)2 }{ h}

=limh012(32+23h+h2)212(3)2h = \lim_{h \rightarrow 0} \frac{ \frac{1}{2}(3^2+2 \cdot 3 \cdot h+h^2)2- \frac{1}{2}(3)2 }{ h}

=limh012(6h+h2)h = \lim_{h \rightarrow 0} \frac{ \frac{1}{2}(6h+h^2) }{ h}

=limh0126+12h = \lim_{h \rightarrow 0} \frac{1}{2} \cdot 6 + \frac{1}{2} h


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

Rate of Change

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


If the distance to an object at time tt is d(f)d(f)  then its:

VELOCITY =d(+)=s(t)=d'(+) =s(t)

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

ACCELERATION =s(f)= s'(f)


An object dropped near Earth travels d(t)=4.9t2 d(t) = 4.9t^2  meters in tt seconds. How far do you need to drop an object so that it hits the ground at 9.8m/s9.8m/s?

Find speed at time tt

d(t)=9.8td'(t) =9.8t

Solve for d(t)=9.8d'(t) =9.8

d(t)=9.8t=9.81d'(t) = 9.8t =9.8 \Rightarrow1

Find the distance traveled at t=1t =1

d(1)=4.9t2=4.9md(1)=4.9t^2 =4.9m

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


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

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

s=2p(1p)=2p2p2 s= 2p(1-p)=2p-2p^2

When does ss change most as a function of pp?

Find dsdp\frac {ds}{dp} 

dsdp=limh0s(p+h)s(p)(p+h)p\frac{ds}{dp} = \lim_{h \rightarrow 0} \frac{s(p+h) -s(p)}{(p+h)-p} 

=limh0s(p+h)s(p)h= \lim_{h \rightarrow 0} \frac {s(p+h)-s(p)}{h}

=limh02(p+h)2(p+h)2][2p2p2]h= \lim_{h \rightarrow 0} \frac {2(p+h)-2(p+h)^2]-[2p-2p^2]}{h} 

=limh02h4ph2h2h= \lim_{h \rightarrow 0} \frac {2h-4ph-2h^2}{h}

=limh024p2h= \lim_{h \rightarrow 0} 2-4p-2h


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


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


Why is this true?

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \rightarrow 0 } \frac{f(x+h)-f(x)}{h} \rightarrow Substitute f(x)=kf(x)=k

=limh0kkh= \lim_{h \rightarrow 0 } \frac{k-k}{h}

ddx(k)=0\frac{d}{dx}(k) =0


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


Why is this true ? \rightarrow  Transformation

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \rightarrow 0} \frac { f(x+h)-f(x)}{h}  \rightarrow Substitute f(x)=cg(x)f(x) = c \cdot g(x)

=limh0cg(x+h)cg(x)h= \lim_{h \rightarrow 0} \frac {c \cdot g(x+h)-c \cdot g(x)}{h}

=climh0g(x+h)g(x)h=c \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h }  \rightarrow Limit Law


ddx[cg(x)]=cddx[g(x)]\frac{d}{dx}[c \cdot g(x)] =c \frac {d}{dx}[g(x)]


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

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

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