Pure 6.2 - Gradients of exponential functions | Edexcel A-Level Mathematics

Find the gradient of $e^{kx}$ and connect that gradient to the current height of the curve. Use the connection to recognise and interpret exponential growth and decay models, including their coefficients, signs and units.

The gradient rule

If a quantity becomes twice as large, must its rate of change double too? For $y=e^{3x}$ , the gradient at height $1$ is $3$ ; at height $2$ , it is $6$ . The rate changes, but its ratio to the current height stays fixed. This is the clue behind exponential models.

You already know

A curve's gradient at a point is the gradient of its tangent there. The derivative $\mathrm{d}y/\mathrm{d}x$ gives the instantaneous rate of change of $y$ with respect to $x$ . Also, $e^0=1$ and $e^x>0$ .

The special feature of the base $e$ is that $y=e^x$ has gradient $e^x$ : its gradient equals its height. Replacing $x$ by $kx$ changes the exponent $k$ times as fast, so an extra factor $k$ appears. A negative $k$ reverses the direction.

Gradient of an exponential

For a constant $k$ ,

\frac{\mathrm{d}}{\mathrm{d}x}(e^x)=e^x, \qquad \frac{\mathrm{d}}{\mathrm{d}x}(e^{kx})=ke^{kx}

This includes $k=0$ : the constant $e^0=1$ has gradient $0$ .
Not in the formula booklet: learn it.

Keep the exponent unchanged. Only multiply the whole exponential by its constant exponent coefficient.

Worked example 1: a growing exponential

Question. Find the gradient function of $y=e^{3x}$ , and the gradient at $x=0$ .

Plan. Multiply by $3$ , then substitute the input into the gradient function.

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=3e^{3x} && \text{exponent coefficient is }3 \\ \left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=0}&=3e^0=3 && \text{substitute after differentiating} \end{aligned}

Why this step? The point's height is $1$ , but its tangent's gradient is $3$ ; these are different quantities when $k\ne1$ .

Answer. Gradient function $3e^{3x}$ ; gradient at zero $3$ .

Check. The function is increasing, consistent with a positive gradient.

Change only the exponent coefficient from $3$ to $-0.4$ . The same rule must now produce a negative gradient.

Worked example 2: a decreasing exponential

Question. Find the gradient function of $z=e^{-0.4x}$ , and the gradient at $x=0$ .

Plan. Keep the exponent $-0.4x$ and multiply by $-0.4$ .

\begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}x}&=-0.4e^{-0.4x} && \text{include the negative coefficient} \\ \left.\frac{\mathrm{d}z}{\mathrm{d}x}\right|_{x=0}&=-0.4e^0=-0.4 && \text{evaluate at zero} \end{aligned}

Why this step? The negative sign describes decay; dropping it would turn the tangent uphill.

Answer. Gradient function $-0.4e^{-0.4x}$ ; gradient at zero $-0.4$ .

Check. A small positive increase in $x$ reduces $e^{-0.4x}$ , consistent with a negative tangent gradient.

The operation is multiplication by $k$ , not multiplication by the whole exponent.

Watch out

Wrong: $\frac{\mathrm{d}}{\mathrm{d}x}(e^{5x})=5x e^{5x}$ . Right: $5e^{5x}$ . Differentiating the linear exponent supplies $5$ , rather than $5x$ .

Change the negative coefficient again and try both stages.

Using the $y$ -value

Sometimes the question gives you the height rather than the input. That is useful information: for $y=e^{kx}$ , the derivative is $ke^{kx}=ky$ . You can use the height directly, without finding $x$ .

Gradient in terms of current height

If $y=e^{kx}$ , with constant $k$ , then

\frac{\mathrm{d}y}{\mathrm{d}x}=ky \qquad\text{and}\qquad \frac{\text{gradient}}{\text{current height}}=k

The ratio is defined because $y>0$ .
Not in the formula booklet: learn it.

This rewriting often removes most of the calculation.

Worked example 3: using a supplied height

Question. A point $P$ on $y=e^{0.7x}$ has $y$ -coordinate $5$ . Find the gradient at $P$ .

Plan. Replace the exponential in the derivative by $y$ .

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=0.7e^{0.7x} && \text{differentiate} \\ &=0.7y && \text{use the curve equation} \\ \text{gradient at }P&=0.7(5)=3.5 && \text{use the given height} \end{aligned}

Why this step? The question already supplies $e^{0.7x}$ through the value of $y$ ; finding $x$ would add work.

Answer. The gradient at $P$ is $3.5$ .

Check. The ratio $3.5/5=0.7$ , as required.

Change only the exponent coefficient from $0.7$ to $-0.4$ , keeping the supplied height $5$ . The sign of the gradient must change.

Worked example 4: decay at the same height

Question. A point $Q$ on $y=e^{-0.4x}$ has $y$ -coordinate $5$ . Find the gradient at $Q$ .

Plan. Rewrite the derivative as $-0.4y$ , then use the supplied height.

\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=-0.4e^{-0.4x}=-0.4y && \text{differentiate and recognise }y \\ \text{gradient at }Q&=-0.4(5)=-2 && \text{substitute the height} \end{aligned}

Why this step? A positive height can have a negative tangent gradient; its sign comes from $k$ .

Answer. The gradient at $Q$ is $-2$ .

Check. The ratio of gradient to height is $-2/5=-0.4$ , and the curve is decreasing.

The sign belongs to the proportionality constant, even though the height is positive.

Key idea

For $y=e^{kx}$ , “gradient equals $k$ times the current height” works for both growth and decay.

Keep that negative proportionality constant and change only the height.

Why exponential models fit proportional rates

Suppose a model has $80$ units initially and an instantaneous growth rate equal to $3\%$ of its current amount per year. At an amount of $80$ , that means $2.4$ units per year; at $160$ , it means $4.8$ . The rate is not constant. Its ratio to the amount is constant.

Rate proportional to the current amount

A quantity $y$ changes at a rate proportional to its current value if $\mathrm{d}y/\mathrm{d}t=ky$ for a fixed constant $k$ . For a positive amount, $k>0$ describes growth and $k<0$ describes decay. A constant rate such as $\mathrm{d}y/\mathrm{d}t=4$ is a different model.

Multiplying an exponential by a constant scales its gradient by the same constant. Thus

\begin{aligned} y=Ae^{kt}&\quad\Rightarrow\quad\frac{\mathrm{d}y}{\mathrm{d}t}=Ake^{kt} && \text{differentiate} \\ &\quad\Rightarrow\quad\frac{\mathrm{d}y}{\mathrm{d}t}=ky && \text{recognise }Ae^{kt}=y \end{aligned}

Here $y(0)=Ae^0=A$ , so $A$ is the initial amount. These model relationships are not in the formula booklet: learn them. For the opening example, $N=80e^{0.03t}$ has derivative $2.4e^{0.03t}=0.03N$ .

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 6.2 - Gradients of exponential functions

The gradient rule

Using the yyy-value

Why exponential models fit proportional rates

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Using the $y$ -value