Chapter 18: Implicit Differentiation

So far, we have covered functions with explicit formulas:

y=f(x)y=f(x)

Sometimes we do not get a formula but instead we get a relation.

x2+y2+sin(xy)=0x^{2}+y^{2}+sin(xy)=0

We cannot solve for yy in this equation


Example

Find dydx\frac{dy}{dx} if x2+y2+sin(xy)=0x^{2}+y^{2}+sin(xy)=0

Assume y=f(x)y=f(x)  for some unknown function

x2+y2+sin(xy)=0x^{2}+y^{2}+sin(xy)=0

x2+[f(x)]2+sin(xf(x)=0x^{2}+[f(x)]^{2}+sin(xf(x)=0

Differentiate the relation

ddx[x2+[f(x)]2+sin(xf(x)]=ddx[0]\frac{d}{dx}[x^{2}+[f(x)]^{2}+sin(xf(x)]=\frac{d}{dx}[0]

2x+2(x)f(x)+cos(xf(x))[f(x)+xf(x)]=02x+2(x)f'(x)+cos(xf(x))[f(x)+xf'(x)]=0

Replace y=f(x)y=f(x) and dydx=f(x)\frac{dy}{dx}=f'(x)

2x+2ydydx+cos(xy)[y+xdydx]=02x+2y \frac{dy}{dx}+cos(xy)[y+x \frac{dy}{dx}]=0

Solve for dydx\frac{dy}{dx}

dydx[2y+xcos(xy)]=2xycos(xy)\frac{dy}{dx}[2y+xcos(xy)]=-2x-ycos(xy)

dydx=2xycos(xy)2y+xcos(xy)\frac{dy}{dx}=-\frac{-2x-ycos(xy)}{2y+xcos(xy)}


Example

Find the slope of the tangent line to

x2+y22=1x^{2}+\frac{y2}{2} =1

At the point (x,y)=(12,32)(x,y)=(\frac{1}{2},\sqrt\frac{3}{2})

Differentiate both the sides

ddx[x2+y22]=ddx[1]\frac{d}{dx}[x2+\frac{y^{2}}{2}]=\frac{d}{dx}[1]

2x+2y2dydx=02x+\frac{2y}{2}\cdot \frac{dy}{dx}=0

2x+ydydx=0dydx=2xy2x+y \frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{-2x}{y}

Evaluate for the sign

dxdy=21232=23\frac{dx}{dy}=\frac{-2\frac{1}{2}}{\sqrt{\frac{3}{2}}}=-\sqrt{\frac{2}{3}}


Discuss

The fundamental law of muscle contraction

(p+a)(v+b)=k(p+a)(v+b)=k 

a,b,ka,b,k - Positive constants

pp - Load

vv - Velocity of contraction

Find dvdP\frac{dv}{dP} and interpret this physically

ddP[(P+a)(v+b)]=ddP[k]\frac{d}{dP}[(P+a)(v+b)]=\frac{d}{dP}[k]

(1+0)(v+b)+(P+a)(dvdP+0)=0(1+0)(v+b)+(P+a)(\frac{dv}{dP}+0)=0

dvdP=v+bP+a=k(P+a)2<0\frac{dv}{dP}= -\frac{v+b}{P+a}= -\frac{k}{(P+a)^{2}}<0



Derivatives and Inverses and Logarithmic Differentiation

Derivatives and Inverses

Recall, the inverse of a function f(x)f(x) is a function g(x)g(x)  such that:

f(g(x())=g(f(x))=xf(g(x())=g(f(x))=x


EXAMPLE

Find the inverse of y=f(x)=2x1y=f(x)=2x_1

Solve for xx in terms of yy

y=2x+1y=2x+1

y1=2xy-1=2x

x=y12x=\frac{y-1}{2}

Switch xx and yy Define g(x)=yg(x)=y

y=x12y=\frac{x-1}{2}   g(x)=x12g(x)=\frac{x-1}{2}  

Check:

f(g(x))f(g(x)) g(f(x))g(f(x))

=f(x12)=f(\frac{x-1}{2}) =g(2x+1)=g(2x+1)

=2(x12)+1=2(\frac{x-1}{2})+1 =(2x+1)12= \frac{(2x+1)-1}{2}

=x1=x= x-1=x =2x2=2= \frac{2x}{2} = 2

Thus, g(x)=x12g(x) = \frac{x-1}{2} is the inverse of f(x)=2x+1f(x)=2x+1.


The notation g(x)=f1(x)g(x)=f^{-1}(x) is very common.


QUESTION

If f(f1(x))=f1(f(x))=xf(f^{-1}(x)) = f^{-1}(f(x)) = x then how are ddx[d1(x)]\frac{d}{dx}[d^{-1}(x)]  and ddx[f(x)]\frac{d}{dx} [f(x)] related? (HINT: Use the chain rule)


FACT

ddx[f1(x)]=1f(f1(x))\frac{d}{dx}[f^{-1}(x)]= \frac{1}{f'(f^{-1}(x))}


DISCUSS

If f(x)=exf(x) = e^x  find ddx[f1(x)]\frac{d}{dx}[f^{-1}(x)]

Recall, f1(x)=ln(x)f^{-1} (x) = \ln(x)

ddx[f1(x)]=1f(f1(x)=1eln(x)=1x\frac{d}{dx}[f^{-1}(x)]= \frac{1}{f'(f^{-1}(x)}=\frac{1}{e^{\ln(x)}}= \frac{1}{x}


KEY FACT

ddx[ln(x)]=1x\frac{d}{dx} [\ln(x)] = \frac{1}{x}


EXAMPLE

Find ddx[2x]\frac{d}{dx}[2^x] and ddx[log2(x)]\frac{d}{dx}[\log_2(x)]

ddx[2x]\frac{d}{dx}[2^x] =ddx[(eln(2))x]= \frac{d}{dx}[(e^{\ln(2)})^x] =ddx[exln(2)]= \frac{d}{dx}[e^{x \ln(2)}]

=d exln(2)dxln(2)=\frac{d \ e^{x \ln(2)}}{dx \ln(2)} \cdot  dxln(2)dx\frac{dx \ln(2)}{dx}

=(exln(2))(ln2)=(e^{x \ln(2)})(\ln 2)

=ln(2)2x= \ln(2) 2^x

ddx[log2(x)]=1ln(2)2log2(x)=1ln(2)x\frac{d}{dx} [\log _2 (x)]= \frac{1}{ln(2)2^{ \log_2(x)}}= \frac{1}{\ln (2) \cdot x}


DISCUSS

Calculate using a calculator

ddx[2x]\frac{d}{dx} [2^x]  and ddx[ex]\frac{d}{dx}[e^x]  at x=0x=0 and compare with ddx[ex]\frac{d}{dx} [e^x]


EXAMPLE

Suppose a population of 100 bacteria doubles in size every six hour. At what rate is the population increasing at 15 hours?

Introduce Notation

P(t)=P(t) = population at time tt in hours

Find a Formula

Initial populatoion =100=100

Number of times we double in tt hours =6=6

P(t)=1002t/6P(t)=100 \cdot 2^{t/6}

Calculate the instant rate of change

P(t)=ddt[1002(t/6)]P'(t)= \frac{d}{dt}[100 \cdot 2 ^{(t/6)}]

=100ddt[2(t/6)]=100 \frac{d}{dt} [2^{(t/6)}]

=100d2(t/6)d(t/6)d(t/6)dt= 100 \frac{d 2 ^{(t/6)}}{d(t/6)} \cdot \frac{d(t/6)}{dt}

=100ln(2)2(t/6)16=100 \cdot \ln(2) 2^{(t/6)} \cdot \frac {1}{6}

Find the required rate

P(15)=100ln(2)2(15/16)16P'(15) = 100 \cdot \ln(2) \cdot 2^{(15/16)} \cdot \frac{1}{6}

22.12\approx 22.12

Thus, 22.1222.12 bacteria/hr are created.



Logarithmic Differentiation

When modelling exponential behaviours, such as population growth, it's important to study ln(f(x))\ln(f(x))


Consider a population of bacteria P(+)P(+) which is given access to food. We see a sharp increase in ln(P(+))\ln (P(+)) which food is available.


QUESTION

How does ddt[ln(P(t))]\frac{d}{dt} [\ln (P(t))]  relate to P(t)P(t)?

ddt[ln(P(t))]=dln(P(t))d P(t)d P(t)dt\frac{d}{dt} [\ln (P(t))] = \frac {d \ln (P(t))}{d \ P(t)} \cdot \frac{d \ P(t)}{dt}

=1P(t)P(t)=P(t)P(t)= \frac{1}{P(t)} \cdot P'(t) = \frac{P'(t)}{P(t)}


EXAMPLE

Calculate f(x)f'(x)  if f(x)=ln(x2+x+1)f(x) = \ln(x^2 + x +1)

f(t)=ddx[x2+x+1]x2+x+1=2x+1x2+x+1f'(t) = \frac{\frac{d}{dx} [x^2+x+1]}{x^2+x+1}= \frac{2x+1}{x^2+x+1}


EXAMPLE

Calculate g(x)g'(x) if g(x)=ln(ex+x2)g(x) =\ln (e^x+x^2)

g(x)=ddx[ex]+x2ex+x2=ex+2xex+x2g'(x) = \frac {\frac{d}{dx}[e^x]+x^2}{e^x+x^2} = \frac{e^x+2x}{e^x+x^2}







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