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

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

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

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

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

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

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

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

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