# Chapter 18: Implicit Differentiation

So far, we have covered functions with explicit formulas:

﻿$y=f(x)$﻿

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

﻿$x^{2}+y^{2}+sin(xy)=0$﻿

We cannot solve for ﻿$y$﻿ in this equation

#### Example

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

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

﻿$x^{2}+y^{2}+sin(xy)=0$﻿

﻿$x^{2}+[f(x)]^{2}+sin(xf(x)=0$﻿

Differentiate the relation

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

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

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

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

Solve for ﻿$\frac{dy}{dx}$﻿

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

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

#### Example

Find the slope of the tangent line to

﻿$x^{2}+\frac{y2}{2} =1$﻿

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

Differentiate both the sides

﻿$\frac{d}{dx}[x2+\frac{y^{2}}{2}]=\frac{d}{dx}$﻿

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

﻿$2x+y \frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{-2x}{y}$﻿

Evaluate for the sign

﻿$\frac{dx}{dy}=\frac{-2\frac{1}{2}}{\sqrt{\frac{3}{2}}}=-\sqrt{\frac{2}{3}}$﻿

### Discuss

The fundamental law of muscle contraction

﻿$(p+a)(v+b)=k$﻿

﻿$a,b,k$﻿ - Positive constants

﻿$p$﻿ - Load

﻿$v$﻿ - Velocity of contraction

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

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

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

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

## Derivatives and Inverses and Logarithmic Differentiation

### Derivatives and Inverses

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

﻿$f(g(x())=g(f(x))=x$﻿

#### EXAMPLE

Find the inverse of ﻿$y=f(x)=2x_1$﻿

Solve for ﻿$x$﻿ in terms of ﻿$y$﻿

﻿$y=2x+1$﻿

﻿$y-1=2x$﻿

﻿$x=\frac{y-1}{2}$﻿

Switch ﻿$x$﻿ and ﻿$y$﻿ Define ﻿$g(x)=y$﻿

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

Check:

﻿$f(g(x))$﻿ ﻿$g(f(x))$﻿

﻿$=f(\frac{x-1}{2})$﻿ ﻿$=g(2x+1)$﻿

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

﻿$= x-1=x$﻿ ﻿$= \frac{2x}{2} = 2$﻿

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

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

#### QUESTION

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

#### FACT

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

#### DISCUSS

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

Recall, ﻿$f^{-1} (x) = \ln(x)$﻿

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

#### KEY FACT

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

#### EXAMPLE

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

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

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

﻿$=(e^{x \ln(2)})(\ln 2)$﻿

﻿$= \ln(2) 2^x$﻿

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

#### DISCUSS

Calculate using a calculator

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

#### 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

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

Find a Formula

Initial populatoion ﻿$=100$﻿

Number of times we double in ﻿$t$﻿ hours ﻿$=6$﻿

﻿$P(t)=100 \cdot 2^{t/6}$﻿

Calculate the instant rate of change

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

﻿$=100 \frac{d}{dt} [2^{(t/6)}]$﻿

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

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

Find the required rate

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

﻿$\approx 22.12$﻿

Thus, ﻿$22.12$﻿ bacteria/hr are created.

#### Logarithmic Differentiation

When modelling exponential behaviours, such as population growth, it's important to study ﻿$\ln(f(x))$﻿ Consider a population of bacteria ﻿$P(+)$﻿ which is given access to food. We see a sharp increase in ﻿$\ln (P(+))$﻿ which food is available.

#### QUESTION

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

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

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

#### EXAMPLE

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

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

#### EXAMPLE

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

﻿$g'(x) = \frac {\frac{d}{dx}[e^x]+x^2}{e^x+x^2} = \frac{e^x+2x}{e^x+x^2}$﻿