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

What you'll learn

How the gradient of an exponential curve is connected to its height.
How to differentiate functions of the form $e^{kx}$ .
How the sign and size of $k$ affect the gradient.
Why exponential functions are suitable models for growth and decay.

Prerequisites

The number $e$

The number $e$ is a mathematical constant with approximate value 2.718. It is the base of the natural exponential function $e^x$ .

The graph of $y=e^x$ :

passes through $(0, 1)$ because $e^0=1$ ;
is always positive;
increases as $x$ increases;
has the $x$ -axis as a horizontal asymptote as $x→−∞x\to-\infty$ .

Definition

Natural exponential function

The natural exponential function is the function $f(x)=e^x$ , where $e$ is the mathematical constant approximately equal to 2.718.

Gradient and derivative

The gradient of a curve at a point is the gradient of the tangent to the curve at that point. It measures the curve's instantaneous rate of change.

The derivative of $y = f (x)$ is written as either $f^{'} (x)$ or $dydx\dfrac{dy}{dx}$ . It gives a formula for the gradient at any point on the curve.

For example, if $dydx=6x\dfrac{dy}{dx}=6x$ , then the gradient at $x = 2$ is 12.

Key Idea

A special exponential function

The function $e^x$ is unique because its derivative is equal to the original function:

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

Therefore, at every point on $y=e^x$ , the gradient is equal to the $y$ -coordinate.

Example

Finding a gradient on the curve

Find the gradient of $y=e^x$ at the point where $x=ln⁡5x=\ln 5$ .

Differentiate the function:
$dydx=ex.\frac{dy}{dx}=e^x.$
Substitute $x=ln⁡5x=\ln 5$ into the derivative:
$dydx∣x=ln⁡5=eln⁡5.\left.\frac{dy}{dx}\right|_{x=\ln 5}=e^{\ln 5}.$
Use $e^{\ln 5}=5$ , so the gradient is 5. Notice that the point also has $y$ -coordinate 5.

Differentiating $e^{kx}$

Consider the function

y=e^{kx},

where $k$ is a constant.

The expression $k x$ is an inner function: it is the expression being used as the exponent. Differentiating a composite function such as this requires the chain rule.

Definition

Chain rule

If $y = f (g (x))$ , then the chain rule states that you differentiate the outer function and multiply by the derivative of the inner function:

dydx=f′(g(x))g′(x).\frac{dy}{dx}=f'(g(x))g'(x).

For $y=e^{kx}$ :

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

Therefore,

dydx=ekx×k=kekx.\frac{dy}{dx}=e^{kx}\times k=ke^{kx}.

Key Idea

The main differentiation rule

For any constant $k$ ,

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

Differentiate the exponent $k x$ to get $k$ , then multiply by the original exponential expression.

Example

Differentiating an exponential function

Differentiate $y=e^{5x}$ and find its gradient at $x=ln⁡2x=\ln 2$ .

The derivative of the exponent $5 x$ is 5, so apply the chain rule:
$dydx=5e5x.\frac{dy}{dx}=5e^{5x}.$
Substitute $x=ln⁡2x=\ln 2$ :
$dydx∣x=ln⁡2=5e5ln⁡2.\left.\frac{dy}{dx}\right|_{x=\ln 2} =5e^{5\ln 2}.$
Use $5\ln 2=\ln(2^5)=\ln 32$ :
$5e5ln⁡2=5eln⁡32=5×32=160.5e^{5\ln 2}=5e^{\ln 32}=5\times 32=160.$
The gradient is 160.

Common Mistake

Forgetting the chain-rule multiplier

The derivative of $e^{5x}$ is $5e^{5x}$ , not just $e^{5x}$ . Always multiply by the derivative of the exponent.

How $k$ affects the gradient

For $y=e^{kx}$ ,

dydx=kekx.\frac{dy}{dx}=ke^{kx}.

Since $e^{kx}$ is always positive, the sign of the gradient is determined entirely by the sign of $k$ :

If $k > 0$ , then $dydx>0\dfrac{dy}{dx}>0$ , so the curve is increasing.
If $k < 0$ , then $dydx<0\dfrac{dy}{dx}<0$ , so the curve is decreasing.
If $k = 0$ , then $y=e^0=1$ , which is a horizontal line with gradient zero.

At $x = 0$ ,

dydx∣x=0=ke0=k.\left.\frac{dy}{dx}\right|_{x=0}=ke^0=k.

Therefore, $k$ itself is the gradient where the curve crosses the $y$ -axis.

The curves below show how different values of $k$ affect the steepness and direction of $y=e^{kx}$ .

$Graphs of y = e^x, y = e^{2x} and y = e^{-x}, with their tangents at x = 0$

Example

Determining where the gradient is a given value

For the curve $y=e^{3x}$ , find the exact value of $x$ where the gradient is 12.

Differentiate the function:
$dydx=3e3x.\frac{dy}{dx}=3e^{3x}.$
Set the derivative equal to the required gradient:
$3e^{3x}=12.$
Divide by 3 and take natural logarithms:
$e3x=4⇒3x=ln⁡4.e^{3x}=4 \quad\Rightarrow\quad 3x=\ln 4.$
Hence,
$x=ln⁡43.x=\frac{\ln 4}{3}.$

Tip

A quick sign check

Because $e^{kx}>0$ , the derivative $ke^{kx}$ must have the same sign as $k$ . If your derivative has the wrong sign, check the chain-rule multiplier.

A constant multiple of $e^{kx}$

Exponential models are often written as

y=Ae^{kx},

where $A$ and $k$ are constants.

The constant multiple rule gives

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

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

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

This equation says that the rate of change of $y$ is proportional to the current value of $y$ .

At $x = 0$ ,

y=Ae^0=A,

so $A$ represents the initial value of the model.

Definition

Exponential model

An exponential model has the form $y=Ae^{kx}$ . The constant $A$ is the value when $x = 0$ , while $k$ controls the continuous proportional rate of growth or decay.

Example

Differentiating a population 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 after 10 years.

At $t = 0$ ,
$P(0)=800e^0=800,$
so the initial population is 800.
Differentiate with respect to $t$ :
$dPdt=800(0.04)e0.04t=32e0.04t.\frac{dP}{dt}=800(0.04)e^{0.04t} =32e^{0.04t}.$
At $t = 10$ ,
$dPdt∣t=10=32e0.4.\left.\frac{dP}{dt}\right|_{t=10} =32e^{0.4}.$
The exact rate is $32e^{0.4}$ people per year, approximately 47.7 people per year.

Why exponential models are suitable

Many real quantities change at a rate that depends on how much is currently present.

For example:

a large population may produce more new individuals than a small population;
a larger investment may earn more interest than a smaller investment;
a radioactive sample with more undecayed nuclei has more nuclei available to decay.

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

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

This is exactly the mathematical condition that the rate of change is proportional to the current quantity.

When $k > 0$ , the model represents exponential growth.
When $k < 0$ , the model represents exponential decay.
The magnitude $∣ k ∣$ controls how quickly the quantity changes.

If $x$ is measured in time, the units of $k$ must be reciprocal time, such as per year, so that the exponent $k x$ has no units.

Common Mistake

Confusing the amount with its rate of change

For $y=Ae^{kx}$ , the amount is $Ae^{kx}$ but the rate of change is $Ake^{kx}$ . They are proportional, but they are not equal unless $k = 1$ .

Common Mistake

Limits of an exponential model

An exponential model assumes that the same proportional relationship continues. Real populations and investments may face changing conditions, so the model may only be reasonable over a limited interval.

Exam technique

In the exam

Identify the exponent and differentiate it before writing the chain-rule multiplier.
For $y=Ae^{kx}$ , use $dydx=Akekx\dfrac{dy}{dx}=Ake^{kx}$ and simplify using $dydx=ky\dfrac{dy}{dx}=ky$ when useful.
Check the sign of your derivative against the context: growth requires $k > 0$ , while decay requires $k < 0$ .
Keep answers involving $e$ or logarithms exact unless the question requests a decimal.

Self review

Check yourself

What is the derivative of $e^{-4x}$ ?
At what value of $x$ does $y=e^{2x}$ have gradient 10?
Why does the equation $dydx=ky\dfrac{dy}{dx}=ky$ make $y=Ae^{kx}$ suitable for modelling growth and decay?

What you'll learn

How the gradient of an exponential curve is connected to its height.
How to differentiate functions of the form $e^{kx}$ .
How the sign and size of $k$ affect the gradient.
Why exponential functions are suitable models for growth and decay.

Prerequisites

The number $e$

The number $e$ is a mathematical constant with approximate value 2.718. It is the base of the natural exponential function $e^x$ .

The graph of $y=e^x$ :

passes through $(0, 1)$ because $e^0=1$ ;
is always positive;
increases as $x$ increases;
has the $x$ -axis as a horizontal asymptote as $x→−∞x\to-\infty$ .

Definition

Natural exponential function

The natural exponential function is the function $f(x)=e^x$ , where $e$ is the mathematical constant approximately equal to 2.718.

Gradient and derivative

The gradient of a curve at a point is the gradient of the tangent to the curve at that point. It measures the curve's instantaneous rate of change.

The derivative of $y = f (x)$ is written as either $f^{'} (x)$ or $dydx\dfrac{dy}{dx}$ . It gives a formula for the gradient at any point on the curve.

For example, if $dydx=6x\dfrac{dy}{dx}=6x$ , then the gradient at $x = 2$ is 12.

Key Idea

A special exponential function

The function $e^x$ is unique because its derivative is equal to the original function:

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

Therefore, at every point on $y=e^x$ , the gradient is equal to the $y$ -coordinate.

Example

Finding a gradient on the curve

Find the gradient of $y=e^x$ at the point where $x=ln⁡5x=\ln 5$ .

Differentiate the function:
$dydx=ex.\frac{dy}{dx}=e^x.$
Substitute $x=ln⁡5x=\ln 5$ into the derivative:
$dydx∣x=ln⁡5=eln⁡5.\left.\frac{dy}{dx}\right|_{x=\ln 5}=e^{\ln 5}.$
Use $e^{\ln 5}=5$ , so the gradient is 5. Notice that the point also has $y$ -coordinate 5.

Differentiating $e^{kx}$

Consider the function

y=e^{kx},

where $k$ is a constant.

The expression $k x$ is an inner function: it is the expression being used as the exponent. Differentiating a composite function such as this requires the chain rule.

Definition

Chain rule

If $y = f (g (x))$ , then the chain rule states that you differentiate the outer function and multiply by the derivative of the inner function:

dydx=f′(g(x))g′(x).\frac{dy}{dx}=f'(g(x))g'(x).

For $y=e^{kx}$ :

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

Therefore,

dydx=ekx×k=kekx.\frac{dy}{dx}=e^{kx}\times k=ke^{kx}.

Key Idea

The main differentiation rule

For any constant $k$ ,

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

Differentiate the exponent $k x$ to get $k$ , then multiply by the original exponential expression.

Example

Differentiating an exponential function

Differentiate $y=e^{5x}$ and find its gradient at $x=ln⁡2x=\ln 2$ .

The derivative of the exponent $5 x$ is 5, so apply the chain rule:
$dydx=5e5x.\frac{dy}{dx}=5e^{5x}.$
Substitute $x=ln⁡2x=\ln 2$ :
$dydx∣x=ln⁡2=5e5ln⁡2.\left.\frac{dy}{dx}\right|_{x=\ln 2} =5e^{5\ln 2}.$
Use $5\ln 2=\ln(2^5)=\ln 32$ :
$5e5ln⁡2=5eln⁡32=5×32=160.5e^{5\ln 2}=5e^{\ln 32}=5\times 32=160.$
The gradient is 160.

Common Mistake

Forgetting the chain-rule multiplier

The derivative of $e^{5x}$ is $5e^{5x}$ , not just $e^{5x}$ . Always multiply by the derivative of the exponent.

How $k$ affects the gradient

For $y=e^{kx}$ ,

dydx=kekx.\frac{dy}{dx}=ke^{kx}.

Since $e^{kx}$ is always positive, the sign of the gradient is determined entirely by the sign of $k$ :

If $k > 0$ , then $dydx>0\dfrac{dy}{dx}>0$ , so the curve is increasing.
If $k < 0$ , then $dydx<0\dfrac{dy}{dx}<0$ , so the curve is decreasing.
If $k = 0$ , then $y=e^0=1$ , which is a horizontal line with gradient zero.

At $x = 0$ ,

dydx∣x=0=ke0=k.\left.\frac{dy}{dx}\right|_{x=0}=ke^0=k.

Therefore, $k$ itself is the gradient where the curve crosses the $y$ -axis.

The curves below show how different values of $k$ affect the steepness and direction of $y=e^{kx}$ .

$Graphs of y = e^x, y = e^{2x} and y = e^{-x}, with their tangents at x = 0$

Example

Determining where the gradient is a given value

For the curve $y=e^{3x}$ , find the exact value of $x$ where the gradient is 12.

Differentiate the function:
$dydx=3e3x.\frac{dy}{dx}=3e^{3x}.$
Set the derivative equal to the required gradient:
$3e^{3x}=12.$
Divide by 3 and take natural logarithms:
$e3x=4⇒3x=ln⁡4.e^{3x}=4 \quad\Rightarrow\quad 3x=\ln 4.$
Hence,
$x=ln⁡43.x=\frac{\ln 4}{3}.$

Tip

A quick sign check

Because $e^{kx}>0$ , the derivative $ke^{kx}$ must have the same sign as $k$ . If your derivative has the wrong sign, check the chain-rule multiplier.

A constant multiple of $e^{kx}$

Exponential models are often written as

y=Ae^{kx},

where $A$ and $k$ are constants.

The constant multiple rule gives

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

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

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

This equation says that the rate of change of $y$ is proportional to the current value of $y$ .

At $x = 0$ ,

y=Ae^0=A,

so $A$ represents the initial value of the model.

Definition

Exponential model

An exponential model has the form $y=Ae^{kx}$ . The constant $A$ is the value when $x = 0$ , while $k$ controls the continuous proportional rate of growth or decay.

Example

Differentiating a population 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 after 10 years.

At $t = 0$ ,
$P(0)=800e^0=800,$
so the initial population is 800.
Differentiate with respect to $t$ :
$dPdt=800(0.04)e0.04t=32e0.04t.\frac{dP}{dt}=800(0.04)e^{0.04t} =32e^{0.04t}.$
At $t = 10$ ,
$dPdt∣t=10=32e0.4.\left.\frac{dP}{dt}\right|_{t=10} =32e^{0.4}.$
The exact rate is $32e^{0.4}$ people per year, approximately 47.7 people per year.

Why exponential models are suitable

Many real quantities change at a rate that depends on how much is currently present.

For example:

a large population may produce more new individuals than a small population;
a larger investment may earn more interest than a smaller investment;
a radioactive sample with more undecayed nuclei has more nuclei available to decay.

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

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

This is exactly the mathematical condition that the rate of change is proportional to the current quantity.

When $k > 0$ , the model represents exponential growth.
When $k < 0$ , the model represents exponential decay.
The magnitude $∣ k ∣$ controls how quickly the quantity changes.

If $x$ is measured in time, the units of $k$ must be reciprocal time, such as per year, so that the exponent $k x$ has no units.

Common Mistake

Confusing the amount with its rate of change

For $y=Ae^{kx}$ , the amount is $Ae^{kx}$ but the rate of change is $Ake^{kx}$ . They are proportional, but they are not equal unless $k = 1$ .

Common Mistake

Limits of an exponential model

Exam technique

In the exam

Identify the exponent and differentiate it before writing the chain-rule multiplier.
For $y=Ae^{kx}$ , use $dydx=Akekx\dfrac{dy}{dx}=Ake^{kx}$ and simplify using $dydx=ky\dfrac{dy}{dx}=ky$ when useful.
Check the sign of your derivative against the context: growth requires $k > 0$ , while decay requires $k < 0$ .
Keep answers involving $e$ or logarithms exact unless the question requests a decimal.

Self review

Check yourself

What is the derivative of $e^{-4x}$ ?
At what value of $x$ does $y=e^{2x}$ have gradient 10?
Why does the equation $dydx=ky\dfrac{dy}{dx}=ky$ make $y=Ae^{kx}$ suitable for modelling growth and decay?

What you'll learn

Prerequisites

The number $e$

Gradient and derivative

Differentiating $e^{kx}$

How $k$ affects the gradient

A constant multiple of $e^{kx}$

Why exponential models are suitable

Review 1.9.2 Gradient of e^kx (A-level only) by teaching Genie

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

What you'll learn

Prerequisites

The number $e$

Gradient and derivative

Differentiating $e^{kx}$

How $k$ affects the gradient

A constant multiple of $e^{kx}$

Why exponential models are suitable

Review 1.9.2 Gradient of e^kx (A-level only) by teaching Genie

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

What you'll learn

Prerequisites

The number eee

Gradient and derivative

Differentiating ekxe^{kx}ekx

How kkk affects the gradient

A constant multiple of ekxe^{kx}ekx

Why exponential models are suitable

Review 1.9.2 Gradient of e^kx (A-level only) by teaching Genie

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

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

What you'll learn

Prerequisites

The number eee

Gradient and derivative

Differentiating ekxe^{kx}ekx

How kkk affects the gradient

A constant multiple of ekxe^{kx}ekx

Why exponential models are suitable

Review 1.9.2 Gradient of e^kx (A-level only) by teaching Genie

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

The number $e$

Differentiating $e^{kx}$

How $k$ affects the gradient

A constant multiple of $e^{kx}$

The number $e$

Differentiating $e^{kx}$

How $k$ affects the gradient

A constant multiple of $e^{kx}$