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

Work through these on your own. Write down the constant $k$ in the exponent before you differentiate, and keep the constant in front of the function.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Differentiate $y = e^{3x}$ .
Differentiate $y = e^{-x}$ .
Differentiate $y = 5e^{2x}$ .
Differentiate $y = 2^x$ .
Differentiate $y = 4^{3x}$ .
Find the gradient of $y = e^{2x}$ at the point where $x = 0$ .
Find the gradient of $y = 3e^{-x}$ at the point where $x = 0$ .
Show that $y = e^{3x}$ satisfies the equation $\dfrac{dy}{dx} = 3y$ .
A population is modelled by $P = 200e^{0.05t}$ , where $t$ is in years. Find $\dfrac{dP}{dt}$ when $t = 0$ .
A mass is modelled by $m = 80e^{-0.2t}$ , where $t$ is in hours. Find $\dfrac{dm}{dt}$ when $t = 0$ .
Find the gradient of $y = e^{x/2}$ at the point where $x = 0$ .
Compare the gradients of $y = e^x$ and $y = e^{2x}$ at $x = 0$ .

Answers and concise solutions

1. The derivative

Answer: The derivative of $e^{kx}$ is $ke^{kx}$ . Here $k = 3$ , so $\dfrac{dy}{dx} = 3e^{3x}$ . The curve climbs three times as fast as $e^x$ at every point.

2. The derivative

Answer: Here $k = -1$ , so $\dfrac{dy}{dx} = -1 \times e^{-x} = -e^{-x}$ . The derivative is negative, which matches the falling curve.

3. The derivative

Answer: The constant $5$ stays in front, and the exponent gives a factor $2$ : $\dfrac{dy}{dx} = 5 \times 2e^{2x} = 10e^{2x}$ .

4. The derivative

Answer: For a positive base $a$ , the derivative of $a^x$ is $a^x \ln a$ . So $\dfrac{dy}{dx} = 2^x \ln 2$ . The factor $\ln 2 \approx 0.693$ makes the curve a little less steep than $e^x$ .

5. The derivative

Answer: For $a^{kx}$ the derivative is $ka^{kx}\ln a$ . Here $a = 4$ and $k = 3$ , so $\dfrac{dy}{dx} = 3 \times 4^{3x} \ln 4$ .

6. The gradient

Answer: The derivative is $2e^{2x}$ . At $x = 0$ this is $2e^0 = 2 \times 1 = 2$ . The point on the curve is $(0, 1)$ , and the tangent there has gradient $2$ .

7. The gradient

Answer: The derivative is $-3e^{-x}$ . At $x = 0$ this is $-3e^0 = -3$ . The negative gradient matches the falling curve.

8. The derivative

Answer: The derivative is $3e^{3x}$ . Since $y = e^{3x}$ , this is exactly $3y$ . The rate of change is three times the current value, which is the defining property of the exponential model.

9. The rate at $t = 0$

Answer: Differentiate: $\dfrac{dP}{dt} = 200 \times 0.05e^{0.05t} = 10e^{0.05t}$ . At $t = 0$ this is $10e^0 = 10$ . The population grows by about $10$ individuals per year at the start.

10. The rate at $t = 0$

Answer: Differentiate: $\dfrac{dm}{dt} = 80 \times (-0.2)e^{-0.2t} = -16e^{-0.2t}$ . At $t = 0$ this is $-16$ . The mass starts by falling at about $16$ units per hour, and the rate shrinks as the mass shrinks.

11. The gradient

Answer: The derivative is $\dfrac{1}{2}e^{x/2}$ . At $x = 0$ this is $\dfrac{1}{2}$ . The curve climbs half as fast as $e^x$ at every point, because $k = \dfrac{1}{2}$ .

12. The difference in gradient

Answer: The gradient of $e^x$ at $x = 0$ is $1$ . The gradient of $e^{2x}$ there is $2$ . So the second curve is twice as steep, and the difference is $1$ . Doubling the constant $k$ doubles the gradient at every point.