1.6.2 Gradient of e^(kx) (A-level only)

What you'll learn

How to differentiate functions of the form $e^{kx}$ .
How the constant $k$ controls both the shape and gradient of the graph.
Why exponential functions model continuous growth and decay.
How to find gradients, tangent equations and unknown model parameters.

Prerequisites

Gradient and derivative

The gradient of a curve measures its steepness at a particular point. It is the gradient of the tangent to the curve at that point.

The derivative of a function gives a formula for this gradient. If $y = f (x)$ , derivative notation includes

dydx,f′(x).\frac{dy}{dx}, \qquad f'(x).

Definition

The natural exponential function

The function $e^x$ is the natural exponential function, where $e$ is the mathematical constant approximately equal to 2.718. Its derivative is equal to the function itself:

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

This is a special property: at every point on the graph of $y=e^x$ , the height of the graph and its gradient have the same numerical value.

For example, at $x = 0$ , the height is $e^0=1$ , so the gradient is also 1. At $x = 2$ , the height is $e^2$ , so the gradient is $e^2$ .

The chain rule

To differentiate $e^{kx}$ , you need the chain rule. This is used when one function is applied to another function, forming a composite function.

For $e^{kx}$ :

the outer function is the exponential function;
the inner function is $k x$ .

The chain rule says to differentiate the outer function and then multiply by the derivative of the inner function.

Since

ddx(kx)=k,\frac{d}{dx}(kx)=k,

we obtain the main result.

Key Idea

Gradient of an exponential function

For any constant $k$ ,

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

Differentiate the exponent $k x$ to get the multiplying factor $k$ . The exponential part remains unchanged.

Example

Differentiating exponential functions

Differentiate (a) $y=e^{4x}$ and (b) $y=e^{-3x}$ .

For $y=e^{4x}$ , the derivative of the exponent $4 x$ is 4, so
$dydx=4e4x.\frac{dy}{dx}=4e^{4x}.$
For $y=e^{-3x}$ , the derivative of the exponent $- 3 x$ is $- 3$ , so
$dydx=−3e−3x.\frac{dy}{dx}=-3e^{-3x}.$
The negative derivative in part (b) shows that $e^{-3x}$ is decreasing for every value of $x$ , because $e^{-3x}$ is always positive.

Common Mistake

Forgetting the chain-rule factor

Do not write $ddx(ekx)=ekx\frac{d}{dx}(e^{kx})=e^{kx}$ . You must multiply by the derivative of the exponent, giving $ke^{kx}$ .

How the value of k affects the graph

The exponential expression $e^{kx}$ is positive for every real value of $x$ . Therefore, the sign of its derivative

ke^{kx}

depends entirely on the sign of $k$ .

When k is positive

If $k > 0$ , then $ke^{kx}>0$ . The gradient is positive everywhere, so the graph is increasing. Larger positive values of $k$ produce faster growth and steeper curves.

When k is negative

If $k < 0$ , then $ke^{kx}<0$ . The gradient is negative everywhere, so the graph is decreasing. This is called exponential decay.

In both cases, the graph passes through $(0, 1)$ because $ek⋅0=e0=1e^{k\cdot 0}=e^0=1$ . The line $y = 0$ is a horizontal asymptote: the curve approaches it in one direction but never reaches it.

The graph below compares exponential growth with exponential decay and shows how the tangent gradients change.

$Graphs of y = e^{2x} and y = e^{-x}, with tangent lines showing positive and negative exponential gradients$

Example

Finding a gradient at a point

Find the gradient of $y=e^{2x}$ at the point where $x=ln⁡3x=\ln 3$ .

Differentiate using the exponential rule:
$dydx=2e2x.\frac{dy}{dx}=2e^{2x}.$
Substitute $x=ln⁡3x=\ln 3$ :
$dydx=2e2ln⁡3.\frac{dy}{dx}=2e^{2\ln 3}.$
Use $2\ln 3=\ln(3^2)=\ln 9$ , followed by $e^{\ln 9}=9$ :
$dydx=2×9=18.\frac{dy}{dx}=2\times 9=18.$

Therefore, the gradient is 18.

Tip

Keep exact values

When the input involves logarithms, use identities such as $e^{\ln a}=a$ and $b\ln a=\ln(a^b)$ . This often produces an exact gradient without a calculator.

Including a constant multiplier

Many exponential functions have the form

y=Ae^{kx},

where $A$ and $k$ are constants. The constant $A$ is often the initial value because

y(0)=Aek⋅0=A.y(0)=Ae^{k\cdot 0}=A.

Using the constant multiple rule,

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

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

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

Definition

Rate of change

A rate of change describes how quickly one quantity changes with respect to another. In an exponential model $y=Ae^{kx}$ , the instantaneous rate of change is $dydx=ky\frac{dy}{dx}=ky$ .

Example

Finding the equation of a tangent

Find the equation of the tangent to $y=5e^{2x}$ at $x = 0$ .

Find the point on the curve:
$y=5e2⋅0=5,y=5e^{2\cdot 0}=5,$
so the tangent passes through $(0, 5)$ .
Differentiate to find the gradient function:
$dydx=10e2x.\frac{dy}{dx}=10e^{2x}.$
At $x = 0$ , the tangent gradient is
$10e^0=10.$
Use the straight-line equation $y-y_1=m(x-x_1)$ :
$y - 5 = 10 (x - 0),$
so the tangent is
$y=10x+5.\boxed{y=10x+5}.$

Why exponential models are useful

An exponential model is suitable when a quantity's rate of change is proportional to the quantity currently present.

If $y=Ae^{kx}$ , then

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

The constant $k$ is the constant of proportionality. This equation says that a larger current amount produces a proportionally larger rate of change.

Key Idea

The modelling connection

Exponential models are suitable when the rate of change at any instant is a constant multiple of the current quantity:

dydx∝y⟹dydx=ky.\frac{dy}{dx}\propto y \quad\Longrightarrow\quad \frac{dy}{dx}=ky.

Positive $k$ models growth, while negative $k$ models decay.

This behaviour occurs in situations such as:

population growth when the birth rate per individual is approximately constant;
compound interest when interest is added continuously;
radioactive decay, where each unstable nucleus has the same chance of decaying;
cooling under suitable conditions, when temperature difference is used as the changing quantity.

A model is a mathematical simplification. In reality, unrestricted population growth cannot continue forever because resources are limited, so an exponential model may only be valid over a stated time interval.

Example

Interpreting an exponential population model

A population is modelled by

P=800e^{0.06t},

where $t$ is measured in years. Find its initial population and instantaneous growth rate after 5 years.

At $t = 0$ ,
$P(0)=800e^0=800,$
so the initial population is 800.
Differentiate with respect to time:
$dPdt=800(0.06)e0.06t=48e0.06t.\frac{dP}{dt}=800(0.06)e^{0.06t}=48e^{0.06t}.$
At $t = 5$ ,
$dPdt=48e0.3≈64.8.\frac{dP}{dt}=48e^{0.3}\approx 64.8.$
The population is therefore growing at approximately 64.8 individuals per year at that instant.
Equivalently, $dPdt=0.06P\frac{dP}{dt}=0.06P$ , so the instantaneous growth rate is always 6% of the current population per year.

Common Mistake

Confusing amount with rate

$P (t)$ gives the population, whereas $dPdt\frac{dP}{dt}$ gives how quickly the population is changing. Their values usually have different meanings and different units.

Finding an unknown value of k

Information about a gradient can be used to determine the parameter $k$ . For $y=Ae^{kx}$ , remember that

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

This form is often faster than differentiating and substituting separately.

Example

Determining the model parameter

A quantity follows the model $y=Ae^{kx}$ . At one instant, $y = 120$ and the quantity is decreasing at a rate of 15 units per hour. Find $k$ .

A decreasing rate of 15 means
$dydx=−15.\frac{dy}{dx}=-15.$
Use the relationship $dydx=ky\frac{dy}{dx}=ky$ and substitute the known values:
$- 15 = 120 k .$
Solve for $k$ :
$k=−15120=−18.k=-\frac{15}{120}=-\frac18.$

The negative value confirms that the model represents exponential decay.

Exam technique

In the exam

Differentiate $e^{kx}$ by keeping the exponential expression unchanged and multiplying by $k$ .
For $y=Ae^{kx}$ , use either $dydx=Akekx\frac{dy}{dx}=Ake^{kx}$ or the quicker relationship $dydx=ky\frac{dy}{dx}=ky$ .
Interpret the sign of $k$ : positive means growth and negative means decay.
Distinguish carefully between the value of the model and its rate of change, and include appropriate units when the variables represent measured quantities.

Self review

Check yourself

What is the derivative of $e^{-5x}$ , and what does its sign tell you about the graph?
If $y=7e^{3x}$ , what are the value of $y$ and the gradient when $x = 0$ ?
Why does the equation $dydx=0.04y\frac{dy}{dx}=0.04y$ suggest that an exponential growth model is appropriate?

What you'll learn

How to differentiate functions of the form $e^{kx}$ .
How the constant $k$ controls both the shape and gradient of the graph.
Why exponential functions model continuous growth and decay.
How to find gradients, tangent equations and unknown model parameters.

Prerequisites

Gradient and derivative

The gradient of a curve measures its steepness at a particular point. It is the gradient of the tangent to the curve at that point.

The derivative of a function gives a formula for this gradient. If $y = f (x)$ , derivative notation includes

dydx,f′(x).\frac{dy}{dx}, \qquad f'(x).

Definition

The natural exponential function

The function $e^x$ is the natural exponential function, where $e$ is the mathematical constant approximately equal to 2.718. Its derivative is equal to the function itself:

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

This is a special property: at every point on the graph of $y=e^x$ , the height of the graph and its gradient have the same numerical value.

For example, at $x = 0$ , the height is $e^0=1$ , so the gradient is also 1. At $x = 2$ , the height is $e^2$ , so the gradient is $e^2$ .

The chain rule

To differentiate $e^{kx}$ , you need the chain rule. This is used when one function is applied to another function, forming a composite function.

For $e^{kx}$ :

the outer function is the exponential function;
the inner function is $k x$ .

The chain rule says to differentiate the outer function and then multiply by the derivative of the inner function.

Since

ddx(kx)=k,\frac{d}{dx}(kx)=k,

we obtain the main result.

Key Idea

Gradient of an exponential function

For any constant $k$ ,

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

Differentiate the exponent $k x$ to get the multiplying factor $k$ . The exponential part remains unchanged.

Example

Differentiating exponential functions

Differentiate (a) $y=e^{4x}$ and (b) $y=e^{-3x}$ .

For $y=e^{4x}$ , the derivative of the exponent $4 x$ is 4, so
$dydx=4e4x.\frac{dy}{dx}=4e^{4x}.$
For $y=e^{-3x}$ , the derivative of the exponent $- 3 x$ is $- 3$ , so
$dydx=−3e−3x.\frac{dy}{dx}=-3e^{-3x}.$
The negative derivative in part (b) shows that $e^{-3x}$ is decreasing for every value of $x$ , because $e^{-3x}$ is always positive.

Common Mistake

Forgetting the chain-rule factor

Do not write $ddx(ekx)=ekx\frac{d}{dx}(e^{kx})=e^{kx}$ . You must multiply by the derivative of the exponent, giving $ke^{kx}$ .

How the value of k affects the graph

The exponential expression $e^{kx}$ is positive for every real value of $x$ . Therefore, the sign of its derivative

ke^{kx}

depends entirely on the sign of $k$ .

When k is positive

If $k > 0$ , then $ke^{kx}>0$ . The gradient is positive everywhere, so the graph is increasing. Larger positive values of $k$ produce faster growth and steeper curves.

When k is negative

If $k < 0$ , then $ke^{kx}<0$ . The gradient is negative everywhere, so the graph is decreasing. This is called exponential decay.

The graph below compares exponential growth with exponential decay and shows how the tangent gradients change.

$Graphs of y = e^{2x} and y = e^{-x}, with tangent lines showing positive and negative exponential gradients$

Example

Finding a gradient at a point

Find the gradient of $y=e^{2x}$ at the point where $x=ln⁡3x=\ln 3$ .

Differentiate using the exponential rule:
$dydx=2e2x.\frac{dy}{dx}=2e^{2x}.$
Substitute $x=ln⁡3x=\ln 3$ :
$dydx=2e2ln⁡3.\frac{dy}{dx}=2e^{2\ln 3}.$
Use $2\ln 3=\ln(3^2)=\ln 9$ , followed by $e^{\ln 9}=9$ :
$dydx=2×9=18.\frac{dy}{dx}=2\times 9=18.$

Therefore, the gradient is 18.

Tip

Keep exact values

When the input involves logarithms, use identities such as $e^{\ln a}=a$ and $b\ln a=\ln(a^b)$ . This often produces an exact gradient without a calculator.

Including a constant multiplier

Many exponential functions have the form

y=Ae^{kx},

where $A$ and $k$ are constants. The constant $A$ is often the initial value because

y(0)=Aek⋅0=A.y(0)=Ae^{k\cdot 0}=A.

Using the constant multiple rule,

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

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

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

Definition

Rate of change

A rate of change describes how quickly one quantity changes with respect to another. In an exponential model $y=Ae^{kx}$ , the instantaneous rate of change is $dydx=ky\frac{dy}{dx}=ky$ .

Example

Finding the equation of a tangent

Find the equation of the tangent to $y=5e^{2x}$ at $x = 0$ .

Find the point on the curve:
$y=5e2⋅0=5,y=5e^{2\cdot 0}=5,$
so the tangent passes through $(0, 5)$ .
Differentiate to find the gradient function:
$dydx=10e2x.\frac{dy}{dx}=10e^{2x}.$
At $x = 0$ , the tangent gradient is
$10e^0=10.$
Use the straight-line equation $y-y_1=m(x-x_1)$ :
$y - 5 = 10 (x - 0),$
so the tangent is
$y=10x+5.\boxed{y=10x+5}.$

Why exponential models are useful

An exponential model is suitable when a quantity's rate of change is proportional to the quantity currently present.

If $y=Ae^{kx}$ , then

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

The constant $k$ is the constant of proportionality. This equation says that a larger current amount produces a proportionally larger rate of change.

Key Idea

The modelling connection

Exponential models are suitable when the rate of change at any instant is a constant multiple of the current quantity:

dydx∝y⟹dydx=ky.\frac{dy}{dx}\propto y \quad\Longrightarrow\quad \frac{dy}{dx}=ky.

Positive $k$ models growth, while negative $k$ models decay.

This behaviour occurs in situations such as:

population growth when the birth rate per individual is approximately constant;
compound interest when interest is added continuously;
radioactive decay, where each unstable nucleus has the same chance of decaying;
cooling under suitable conditions, when temperature difference is used as the changing quantity.

Example

Interpreting an exponential population model

A population is modelled by

P=800e^{0.06t},

where $t$ is measured in years. Find its initial population and instantaneous growth rate after 5 years.

At $t = 0$ ,
$P(0)=800e^0=800,$
so the initial population is 800.
Differentiate with respect to time:
$dPdt=800(0.06)e0.06t=48e0.06t.\frac{dP}{dt}=800(0.06)e^{0.06t}=48e^{0.06t}.$
At $t = 5$ ,
$dPdt=48e0.3≈64.8.\frac{dP}{dt}=48e^{0.3}\approx 64.8.$
The population is therefore growing at approximately 64.8 individuals per year at that instant.
Equivalently, $dPdt=0.06P\frac{dP}{dt}=0.06P$ , so the instantaneous growth rate is always 6% of the current population per year.

Common Mistake

Confusing amount with rate

$P (t)$ gives the population, whereas $dPdt\frac{dP}{dt}$ gives how quickly the population is changing. Their values usually have different meanings and different units.

Finding an unknown value of k

Information about a gradient can be used to determine the parameter $k$ . For $y=Ae^{kx}$ , remember that

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

This form is often faster than differentiating and substituting separately.

Example

Determining the model parameter

A quantity follows the model $y=Ae^{kx}$ . At one instant, $y = 120$ and the quantity is decreasing at a rate of 15 units per hour. Find $k$ .

A decreasing rate of 15 means
$dydx=−15.\frac{dy}{dx}=-15.$
Use the relationship $dydx=ky\frac{dy}{dx}=ky$ and substitute the known values:
$- 15 = 120 k .$
Solve for $k$ :
$k=−15120=−18.k=-\frac{15}{120}=-\frac18.$

The negative value confirms that the model represents exponential decay.

Exam technique

In the exam

Differentiate $e^{kx}$ by keeping the exponential expression unchanged and multiplying by $k$ .
For $y=Ae^{kx}$ , use either $dydx=Akekx\frac{dy}{dx}=Ake^{kx}$ or the quicker relationship $dydx=ky\frac{dy}{dx}=ky$ .
Interpret the sign of $k$ : positive means growth and negative means decay.
Distinguish carefully between the value of the model and its rate of change, and include appropriate units when the variables represent measured quantities.

Self review

Check yourself

What is the derivative of $e^{-5x}$ , and what does its sign tell you about the graph?
If $y=7e^{3x}$ , what are the value of $y$ and the gradient when $x = 0$ ?
Why does the equation $dydx=0.04y\frac{dy}{dx}=0.04y$ suggest that an exponential growth model is appropriate?

What you'll learn

Prerequisites

Gradient and derivative

The chain rule

How the value of k affects the graph

When k is positive

When k is negative

Including a constant multiplier

Why exponential models are useful

Finding an unknown value of k

Review 1.6.2 Gradient of e^(kx) (A-level only) by teaching Genie

1.6.2 Gradient of e^(kx) (A-level only)

What you'll learn

Prerequisites

Gradient and derivative

The chain rule

How the value of k affects the graph

When k is positive

When k is negative

Including a constant multiplier

Why exponential models are useful

Finding an unknown value of k

Review 1.6.2 Gradient of e^(kx) (A-level only) by teaching Genie

1.6.2 Gradient of e^(kx) (A-level only)

What you'll learn

Prerequisites

Gradient and derivative

The chain rule

How the value of k affects the graph

When k is positive

When k is negative

Including a constant multiplier

Why exponential models are useful

Finding an unknown value of k

Review 1.6.2 Gradient of e^(kx) (A-level only) by teaching Genie

1.6.2 Gradient of e^(kx) (A-level only) Revision Guide

1.6.2 Gradient of e^(kx) (A-level only)

What you'll learn

Prerequisites

Gradient and derivative

The chain rule

How the value of k affects the graph

When k is positive

When k is negative

Including a constant multiplier

Why exponential models are useful

Finding an unknown value of k

Review 1.6.2 Gradient of e^(kx) (A-level only) by teaching Genie

1.6.2 Gradient of e^(kx) (A-level only) Revision Guide