6.2 The gradient of e^kx and the exponential model — Learn

Prerequisites: the graphs of $a^x$ and $e^x$ from Section 6.1; the gradient of a straight line from Chapter 3; the idea that the gradient of a curve at a point is the gradient of the tangent there. Chapter 7 develops that idea fully.

By the end of this section you can: differentiate $e^{kx}$; differentiate $a^{kx}$ for a positive base $a$; find the gradient of an exponential curve at a given point; explain why an exponential model suits a quantity whose rate of change is proportional to its size; and read a rate of change from a growth or decay model.

1. The derivative of e^{kx}

The derivative of a function gives its gradient. For the exponential function the result is remarkably simple:

The derivative has the same shape as the original function, and the constant $k$ appears in front. When $k = 1$ the derivative of $e^x$ is $e^x$. That is the special property of the base $e$: at every point, the gradient equals the value of the function. A curve that is climbing steeply has a large gradient. The larger the value of $e^{kx}$, the larger the gradient.

For a positive base $a$, write $a = e^{\ln a}$ and use the result above. Then

The extra factor $\ln a$ appears because the base $a$ is not $e$. Section 6.3 explains the logarithm $\ln a$ in full; for now you can evaluate it on a calculator.

Example 1. Differentiate $y = e^{4x}$.

Here $k = 4$, so

Example 2. Differentiate $y = 3e^{-2x}$.

The constant $3$ stays in front, and $k = -2$:

The derivative is negative for every $x$, which matches the falling curve.

Example 3. Differentiate $y = 2^{3x}$.

Here $a = 2$ and $k = 3$, so

Example 4. Find the gradient of $y = e^{2x}$ at the point where $x = 0$.

Differentiate first: $\dfrac{dy}{dx} = 2e^{2x}$. At $x = 0$,

The curve passes through $(0, 1)$, and the tangent there has gradient $2$.

2. The exponential model

Many quantities change at a rate proportional to their current size. A population of bacteria doubles when there is enough food, and the more bacteria there are, the faster the number grows. A radioactive sample decays, and the more atoms there are, the more decay each second. In each case the rate of change is a constant multiple of the amount:

The function that satisfies this equation is the exponential

where $A$ is the value at $t = 0$. Differentiate and check:

• When $k > 0$ the quantity grows. The larger $k$ is, the faster it grows.
• When $k < 0$ the quantity decays. The larger $|k|$ is, the faster it falls.
• The constant $A$ is the starting value, because $e^0 = 1$.

This is why the exponential model suits so many applications. The assumption "the rate is proportional to the amount" is a single, testable statement about the process. It also explains the shape: a growing quantity grows faster as it gets larger, so the curve steepens.

Example 5. A colony of bacteria has $900$ cells at $t = 0$ and grows with constant $k = 0.4$ per hour. Write the model and find the growth rate at $t = 0$.

The model is $N = 900e^{0.4t}$. Differentiate:

At $t = 0$ the rate is $360e^0 = 360$ cells per hour.

Progress check

Return to the cooling coffee in the chapter problem. Its excess temperature is

1. Differentiate $E$ with respect to $t$.
2. Find the value of $\dfrac{dE}{dt}$ at $t = 0$.
3. Explain why the negative sign is expected, and why the coffee cools more slowly as time goes on.

Section 6.4 finds the value of $k$ from the measurement, and then the time for the coffee to cool.

Watch out. These mistakes are common.

Forgetting the constant in the exponent. The derivative of $e^{4x}$ is $4e^{4x}$, not $e^{4x}$. The factor $4$ is what makes a steeper curve have a larger gradient.

Losing the minus sign. The derivative of $e^{-2x}$ is $-2e^{-2x}$. Check the sign against the graph: if the curve is falling, the gradient must be negative.

Dropping $\ln a$ for a base other than $e$. The derivative of $2^{3x}$ is $3 \times 2^{3x}\ln 2$, not $3 \times 2^{3x}$. The factor $\ln 2$ is roughly $0.693$, so the curve is a little less steep than $e^{3x}$.

Mixing up $e^{kx}$ and $x^k$. The first is an exponential and its derivative keeps an exponential. The second is a power, and Chapter 7 shows that its derivative is $kx^{k-1}$. The two rules are different.

You should now be able to

• differentiate $e^{kx}$, keeping the constant $k$ in front;
• differentiate $a^{kx}$ for a positive base $a$, including the factor $\ln a$;
• find the gradient of an exponential curve at a given point;
• write down the model $y = Ae^{kt}$ and say what $A$ and $k$ mean;
• explain why a rate proportional to the amount gives an exponential model;
• find a rate of change from a growth or decay model.