1.6.3 Exponentials and logarithms

What you'll learn

Differentiate functions of the form $e^{kx}$ .
Recognise when a rate of change is proportional to the current value.
Use exponential models for continuous growth and decay.
Interpret the constants in a model of the form $y=Ae^{kx}$ .

Prerequisites: indices and gradients

The index laws allow you to work with exponential expressions. In particular,

eaeb=ea+b,eaeb=ea−b,(ea)b=eab.e^a e^b=e^{a+b}, \qquad \frac{e^a}{e^b}=e^{a-b}, \qquad \left(e^a\right)^b=e^{ab}.

The gradient of a curve measures its instantaneous rate of change. For a function $y = f (x)$ , its gradient is written as

dydx=f′(x).\frac{dy}{dx}=f'(x).

A positive derivative means that $y$ is increasing, while a negative derivative means that $y$ is decreasing.

The number $e$

The number $e$ is an irrational constant with approximate value 2.718. It is the natural base for exponential functions involving continuous change.

The function $y=e^x$ has a particularly important property:

ddx(ex)=ex.\frac{d}{dx}\left(e^x\right)=e^x.

This says that the gradient of $y=e^x$ at every point is equal to the current value of $y$ .

Key Idea

The special property of the exponential function

For $y=e^x$ , the function and its derivative are identical. The larger the value of $y$ , the faster it increases.

Example

Finding a gradient on an exponential curve

Find the gradient of $y=e^x$ at $x = 2$ .

Differentiate the function:
$dydx=ex.\frac{dy}{dx}=e^x.$
Substitute $x = 2$ into the derivative:
$dydx∣x=2=e2.\left.\frac{dy}{dx}\right|_{x=2}=e^2.$
Therefore, the exact gradient is $e^2$ . As a decimal, this is approximately 7.39.

Differentiating $e^{kx}$

Suppose that

y=e^{kx},

where $k$ is a constant. The exponent is now the function $k x$ , so the chain rule is needed.

Definition

Chain rule

If $y$ is a function of $u$ , and $u$ is a function of $x$ , then

dydx=dydududx.\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.

Let $u = k x$ . Then $y=e^u$ , so

dydu=euanddudx=k.\frac{dy}{du}=e^u \qquad\text{and}\qquad \frac{du}{dx}=k.

Therefore,

dydx=eu⋅k=kekx.\frac{dy}{dx}=e^u\cdot k=ke^{kx}.

Key Idea

Derivative of an exponential

For any constant $k$ ,

ddx(ekx)=kekx.\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}.

Differentiate the exponent $k x$ and multiply by its derivative, $k$ .

Example

Differentiating an exponential function

Differentiate $y=5e^{-3x}$ .

The exponent is $- 3 x$ , whose derivative is $- 3$ .
Keep the exponential function unchanged and multiply by $- 3$ :
$dydx=5(−3)e−3x.\frac{dy}{dx}=5\left(-3\right)e^{-3x}.$
Simplify:
$dydx=−15e−3x.\frac{dy}{dx}=-15e^{-3x}.$

The derivative is negative for every value of $x$ , so the function is always decreasing.

Common Mistake

Forgetting the chain-rule factor

The derivative of $e^{kx}$ is not just $e^{kx}$ . You must multiply by the derivative of the exponent, giving $ke^{kx}$ .

Exponential models

An exponential model represents a quantity using a function such as

y=Ae^{kx},

where $A$ and $k$ are constants.

At $x = 0$ ,

y=Aek⋅0=Ae0=A.y=Ae^{k\cdot 0}=Ae^0=A.

Therefore, $A$ is the initial value of the quantity.

Differentiating gives

dydx=Akekx.\frac{dy}{dx}=Ake^{kx}.

Since $y=Ae^{kx}$ , this can be rewritten as

dydx=ky.\frac{dy}{dx}=ky.

The rate of change is therefore proportional to the current value of $y`.

Definition

Proportional rate of change

The rate of change of $y$ is proportional to $y$ when

dydx=ky\frac{dy}{dx}=ky

for some constant of proportionality $k$ .

This is why exponential models occur naturally in applications. If there is more of a quantity, and each unit contributes independently to the change, then a larger quantity produces a larger rate of change.

Examples include:

population growth when each individual has the same average reproductive rate;
compound interest added continuously;
radioactive decay, where each unstable nucleus has the same probability of decaying;
cooling when the rate depends on the temperature difference from the surroundings.

Growth and decay

The sign of $k$ determines the behaviour of $y=Ae^{kx}$ , assuming $A > 0$ .

If $k > 0$ , then $dydx>0\frac{dy}{dx}>0$ , so the model represents exponential growth.
If $k < 0$ , then $dydx<0\frac{dy}{dx}<0$ , so the model represents exponential decay.
If $k = 0$ , then $y = A$ , so there is no growth or decay.

Tip

Checking the sign

In a decay model, the exponent is usually negative. Because $e^{kx}$ is always positive, the sign of the derivative is controlled by the sign of $k$ .

Example

Interpreting an exponential growth model

A population is modelled by

P=800e^{0.04t},

where $t$ is measured in years. Find the initial population and its rate of growth when $t = 5$ .

At $t = 0$ ,
$P=800e^0=800,$
so the initial population is 800.
Differentiate with respect to $t$ :
$dPdt=800(0.04)e0.04t=32e0.04t.\frac{dP}{dt}=800\left(0.04\right)e^{0.04t} =32e^{0.04t}.$
At $t = 5$ ,
$dPdt=32e0.2≈39.1.\frac{dP}{dt}=32e^{0.2}\approx39.1.$

The model predicts that after 5 years, the population is increasing at approximately 39 individuals per year.

Notice that

dPdt=0.04P.\frac{dP}{dt}=0.04P.

This means the population's instantaneous rate of growth is 0.04 times its current size. Equivalently, it is growing continuously at a rate of 4% per year.

Common Mistake

Confusing the value with its rate of change

In the model $P=800e^{0.04t}$ , $P$ is the population, whereas $dPdt\frac{dP}{dt}$ is the number of individuals added per year. They represent different quantities.

Forming a model from a proportional rate

A statement such as “the rate of change of $y$ is proportional to $y$ ” translates into

dydx=ky.\frac{dy}{dx}=ky.

Functions of the form $y=Ae^{kx}$ satisfy this equation, because differentiating them gives $dydx=ky\frac{dy}{dx}=ky$ .

An initial condition, such as the value of $y$ when $x = 0$ , can then be used to find $A$ .

Example

Constructing a decay model

A substance initially has mass 120 g. Its mass decreases at a rate proportional to its current mass, with constant of proportionality $- 0.08$ . Form an exponential model for its mass $M$ after $t$ hours.

Translate the proportional-rate statement into a differential equation:
$dMdt=−0.08M.\frac{dM}{dt}=-0.08M.$
A function satisfying this equation has the form
$M=Ae^{-0.08t}.$
Use the initial condition $M = 120$ when $t = 0$ :
$120=Ae^0=A.$
Therefore, the model is
$M=120e^{-0.08t}.$

Common Mistake

Check what is proportional

Exponential modelling applies when the rate of change is proportional to the quantity itself. If the rate is constant, the appropriate model is linear rather than exponential.

Exam technique

In the exam

Translate “rate of change” into a derivative such as $dydx\frac{dy}{dx}$ , then translate “proportional to $y$ ” into $dydx=ky\frac{dy}{dx}=ky$ .
When differentiating $e^{kx}$ , multiply by $k$ and keep the exponential factor unchanged.
Use the initial condition carefully to find $A$ , and interpret the sign of $k$ to decide whether the model describes growth or decay.
Give an exact exponential value unless the question requests a decimal, and include appropriate units when interpreting a rate.

Self review

Check yourself

What is the derivative of $7e^{-2x}$ ?
How can you tell from $y=Ae^{kx}$ whether the model represents growth or decay?
A population satisfies $dPdt=0.03P\frac{dP}{dt}=0.03P$ and initially contains 500 individuals. What exponential model represents the population?

What you'll learn

Differentiate functions of the form $e^{kx}$ .
Recognise when a rate of change is proportional to the current value.
Use exponential models for continuous growth and decay.
Interpret the constants in a model of the form $y=Ae^{kx}$ .

Prerequisites: indices and gradients

The index laws allow you to work with exponential expressions. In particular,

eaeb=ea+b,eaeb=ea−b,(ea)b=eab.e^a e^b=e^{a+b}, \qquad \frac{e^a}{e^b}=e^{a-b}, \qquad \left(e^a\right)^b=e^{ab}.

The gradient of a curve measures its instantaneous rate of change. For a function $y = f (x)$ , its gradient is written as

dydx=f′(x).\frac{dy}{dx}=f'(x).

A positive derivative means that $y$ is increasing, while a negative derivative means that $y$ is decreasing.

The number $e$

The number $e$ is an irrational constant with approximate value 2.718. It is the natural base for exponential functions involving continuous change.

The function $y=e^x$ has a particularly important property:

ddx(ex)=ex.\frac{d}{dx}\left(e^x\right)=e^x.

This says that the gradient of $y=e^x$ at every point is equal to the current value of $y$ .

Key Idea

The special property of the exponential function

For $y=e^x$ , the function and its derivative are identical. The larger the value of $y$ , the faster it increases.

Example

Finding a gradient on an exponential curve

Find the gradient of $y=e^x$ at $x = 2$ .

Differentiate the function:
$dydx=ex.\frac{dy}{dx}=e^x.$
Substitute $x = 2$ into the derivative:
$dydx∣x=2=e2.\left.\frac{dy}{dx}\right|_{x=2}=e^2.$
Therefore, the exact gradient is $e^2$ . As a decimal, this is approximately 7.39.

Differentiating $e^{kx}$

Suppose that

y=e^{kx},

where $k$ is a constant. The exponent is now the function $k x$ , so the chain rule is needed.

Definition

Chain rule

If $y$ is a function of $u$ , and $u$ is a function of $x$ , then

dydx=dydududx.\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.

Let $u = k x$ . Then $y=e^u$ , so

dydu=euanddudx=k.\frac{dy}{du}=e^u \qquad\text{and}\qquad \frac{du}{dx}=k.

Therefore,

dydx=eu⋅k=kekx.\frac{dy}{dx}=e^u\cdot k=ke^{kx}.

Key Idea

Derivative of an exponential

For any constant $k$ ,

ddx(ekx)=kekx.\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}.

Differentiate the exponent $k x$ and multiply by its derivative, $k$ .

Example

Differentiating an exponential function

Differentiate $y=5e^{-3x}$ .

The exponent is $- 3 x$ , whose derivative is $- 3$ .
Keep the exponential function unchanged and multiply by $- 3$ :
$dydx=5(−3)e−3x.\frac{dy}{dx}=5\left(-3\right)e^{-3x}.$
Simplify:
$dydx=−15e−3x.\frac{dy}{dx}=-15e^{-3x}.$

The derivative is negative for every value of $x$ , so the function is always decreasing.

Common Mistake

Forgetting the chain-rule factor

The derivative of $e^{kx}$ is not just $e^{kx}$ . You must multiply by the derivative of the exponent, giving $ke^{kx}$ .

Exponential models

An exponential model represents a quantity using a function such as

y=Ae^{kx},

where $A$ and $k$ are constants.

At $x = 0$ ,

y=Aek⋅0=Ae0=A.y=Ae^{k\cdot 0}=Ae^0=A.

Therefore, $A$ is the initial value of the quantity.

Differentiating gives

dydx=Akekx.\frac{dy}{dx}=Ake^{kx}.

Since $y=Ae^{kx}$ , this can be rewritten as

dydx=ky.\frac{dy}{dx}=ky.

The rate of change is therefore proportional to the current value of $y`.

Definition

Proportional rate of change

The rate of change of $y$ is proportional to $y$ when

dydx=ky\frac{dy}{dx}=ky

for some constant of proportionality $k$ .

Examples include:

population growth when each individual has the same average reproductive rate;
compound interest added continuously;
radioactive decay, where each unstable nucleus has the same probability of decaying;
cooling when the rate depends on the temperature difference from the surroundings.

Growth and decay

The sign of $k$ determines the behaviour of $y=Ae^{kx}$ , assuming $A > 0$ .

If $k > 0$ , then $dydx>0\frac{dy}{dx}>0$ , so the model represents exponential growth.
If $k < 0$ , then $dydx<0\frac{dy}{dx}<0$ , so the model represents exponential decay.
If $k = 0$ , then $y = A$ , so there is no growth or decay.

Tip

Checking the sign

In a decay model, the exponent is usually negative. Because $e^{kx}$ is always positive, the sign of the derivative is controlled by the sign of $k$ .

Example

Interpreting an exponential growth model

A population is modelled by

P=800e^{0.04t},

where $t$ is measured in years. Find the initial population and its rate of growth when $t = 5$ .

At $t = 0$ ,
$P=800e^0=800,$
so the initial population is 800.
Differentiate with respect to $t$ :
$dPdt=800(0.04)e0.04t=32e0.04t.\frac{dP}{dt}=800\left(0.04\right)e^{0.04t} =32e^{0.04t}.$
At $t = 5$ ,
$dPdt=32e0.2≈39.1.\frac{dP}{dt}=32e^{0.2}\approx39.1.$

The model predicts that after 5 years, the population is increasing at approximately 39 individuals per year.

Notice that

dPdt=0.04P.\frac{dP}{dt}=0.04P.

This means the population's instantaneous rate of growth is 0.04 times its current size. Equivalently, it is growing continuously at a rate of 4% per year.

Common Mistake

Confusing the value with its rate of change

In the model $P=800e^{0.04t}$ , $P$ is the population, whereas $dPdt\frac{dP}{dt}$ is the number of individuals added per year. They represent different quantities.

Forming a model from a proportional rate

A statement such as “the rate of change of $y$ is proportional to $y$ ” translates into

dydx=ky.\frac{dy}{dx}=ky.

Functions of the form $y=Ae^{kx}$ satisfy this equation, because differentiating them gives $dydx=ky\frac{dy}{dx}=ky$ .

An initial condition, such as the value of $y$ when $x = 0$ , can then be used to find $A$ .

Example

Constructing a decay model

Translate the proportional-rate statement into a differential equation:
$dMdt=−0.08M.\frac{dM}{dt}=-0.08M.$
A function satisfying this equation has the form
$M=Ae^{-0.08t}.$
Use the initial condition $M = 120$ when $t = 0$ :
$120=Ae^0=A.$
Therefore, the model is
$M=120e^{-0.08t}.$

Common Mistake

Check what is proportional

Exponential modelling applies when the rate of change is proportional to the quantity itself. If the rate is constant, the appropriate model is linear rather than exponential.

Exam technique

In the exam

Translate “rate of change” into a derivative such as $dydx\frac{dy}{dx}$ , then translate “proportional to $y$ ” into $dydx=ky\frac{dy}{dx}=ky$ .
When differentiating $e^{kx}$ , multiply by $k$ and keep the exponential factor unchanged.
Use the initial condition carefully to find $A$ , and interpret the sign of $k$ to decide whether the model describes growth or decay.
Give an exact exponential value unless the question requests a decimal, and include appropriate units when interpreting a rate.

Self review

Check yourself

What is the derivative of $7e^{-2x}$ ?
How can you tell from $y=Ae^{kx}$ whether the model represents growth or decay?
A population satisfies $dPdt=0.03P\frac{dP}{dt}=0.03P$ and initially contains 500 individuals. What exponential model represents the population?

What you'll learn

Prerequisites: indices and gradients

The number $e$

Differentiating $e^{kx}$

Exponential models

Growth and decay

Forming a model from a proportional rate

Review 1.6.3 Exponentials and logarithms by teaching Genie

1.6.3 Exponentials and logarithms

What you'll learn

Prerequisites: indices and gradients

The number $e$

Differentiating $e^{kx}$

Exponential models

Growth and decay

Forming a model from a proportional rate

Review 1.6.3 Exponentials and logarithms by teaching Genie

1.6.3 Exponentials and logarithms

What you'll learn

Prerequisites: indices and gradients

The number eee

Differentiating ekxe^{kx}ekx

Exponential models

Growth and decay

Forming a model from a proportional rate

Review 1.6.3 Exponentials and logarithms by teaching Genie

1.6.3 Exponentials and logarithms Revision Guide

1.6.3 Exponentials and logarithms

What you'll learn

Prerequisites: indices and gradients

The number eee

Differentiating ekxe^{kx}ekx

Exponential models

Growth and decay

Forming a model from a proportional rate

Review 1.6.3 Exponentials and logarithms by teaching Genie

1.6.3 Exponentials and logarithms Revision Guide

The number $e$

Differentiating $e^{kx}$

The number $e$

Differentiating $e^{kx}$