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The special property of \(\mathrm{e}^x\): the gradient of \(y = \mathrm{e}^{kx}\) is \(k\mathrm{e}^{kx}\), so the gradient is proportional to the value of the function. This is why exponential models fit anything that grows or decays at a rate proportional to its size. It comes up in gradient questions and when you are asked to explain a model.
If \(y = A\mathrm{e}^{kx}\), then \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = ky\)
The gradient of \(y = \mathrm{e}^x\) at \((0, 1)\) is 1
Rate of change proportional to the amount \(\Leftrightarrow\) exponential model
Notes
The gradient of \(\mathrm{e}^x\)
diagram
For \(y = \mathrm{e}^x\), the gradient at every point equals the y-coordinate: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x\)
So at \((0, 1)\) the gradient is 1, and at \((2, \mathrm{e}^2)\) the gradient is \(\mathrm{e}^2\)
At every point of \(y = \mathrm{e}^x\) the gradient equals the y-coordinate.
\(\mathrm{e}\) is the only base for which this is true. For any other \(y = a^x\) the gradient is a constant multiple of \(a^x\) (the constant is \(\ln a\)).
Differentiating \(\mathrm{e}^{kx}\)
\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{kx}\right) = k\mathrm{e}^{kx}\) for any constant \(k\): the power stays the same and \(k\) comes down in front.
Constant multiples stay: \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(A\mathrm{e}^{kx}\right) = Ak\mathrm{e}^{kx}\). For example, \(y = 5\mathrm{e}^{-2x}\) gives \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -10\mathrm{e}^{-2x}\)
If \(k > 0\) the gradient is positive and increases as \(x\) increases. If \(k \lt 0\) the gradient is negative and gets closer to 0.
Since \(A\mathrm{e}^{kx} = y\), the result can be written \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = ky\): the gradient is proportional to \(y\)
Why exponential models are so useful
Many quantities change at a rate proportional to their size: a population (more animals, more births), the number of undecayed radioactive nuclei, money earning continuous interest, the amount of a drug in the blood.
If \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = kN\), then \(N = N_0\mathrm{e}^{kt}\) fits, because differentiating \(N_0\mathrm{e}^{kt}\) gives \(kN_0\mathrm{e}^{kt} = kN\)
\(k > 0\) gives growth that gets faster and faster; \(k \lt 0\) gives decay that gets slower and slower.
In an 'explain' question, say that the rate of change is proportional to the amount, and put it in context.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find the exact gradient at a point
2 to 3 marksGrade D
Differentiate: multiply by the number in front of \(x\) in the power.
Substitute the x-value into \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), not into \(y\)
Leave the answer in terms of \(\mathrm{e}\) if it asks for an exact value.
Example. The curve \(C\) has equation \(y = 4\mathrm{e}^{0.5x}\). Find the exact gradient of \(C\) at the point where \(x = 2\)
Show a rate is proportional to the amount, and explain
2 to 3 marksGrade A
Differentiate the model.
Rewrite the derivative in terms of the original variable, e.g. \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = 0.3N\)
Interpret in context: the rate of increase is proportional to the current amount.
Example. The number of bacteria, \(N\), in a culture \(t\) hours after the start of an experiment is modelled by \(N = 200\mathrm{e}^{0.3t}\). (a) Show that \(\dfrac{\mathrm{d}N}{\mathrm{d}t} = 0.3N\). (b) Explain what this tells you about how the number of bacteria grows.
(b) The rate at which the number of bacteria increases is proportional to the number of bacteria present, so the more bacteria there are, the faster the number grows
B1
Don’t lose marks
Using the power rule on \(\mathrm{e}^x\), e.g. writing \(x\mathrm{e}^{x-1}\). The power rule is only for \(x^n\)
Leaving out the \(k\): the derivative of \(\mathrm{e}^{3x}\) is \(3\mathrm{e}^{3x}\), not \(\mathrm{e}^{3x}\)
Sign slips with negative powers: the derivative of \(\mathrm{e}^{-x}\) is \(-\mathrm{e}^{-x}\)
Substituting into \(y\) instead of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) when asked for a gradient.
Giving 5.44 when the question says 'exact': the answer is \(2\mathrm{e}\)
More tips
Memory tricks
Bring the \(k\) down in front: \(\mathrm{e}^{kx} \to k\mathrm{e}^{kx}\). The power itself never changes.
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = ky\) is the fingerprint of an exponential: if a rate is proportional to the amount, think \(\mathrm{e}\)
Quick check: the tangent to \(y = \mathrm{e}^x\) at \((0, 1)\) is \(y = x + 1\), with gradient 1.
Exam technique
'Exact' means keep \(\mathrm{e}\) and ln in the answer; do not round.
In 'explain' questions use the phrase 'the rate of change (or rate of increase) is proportional to' and name the quantity in the context.
'Show that' answers need the step where you replace \(200\mathrm{e}^{0.3t}\) by \(N\); do not jump straight to the result.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
State the gradient function of \(\mathrm{e}^x\)If \(y = \mathrm{e}^x\) then \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x\): the gradient at every point equals the y-coordinate.
Grade D
Differentiate \(\mathrm{e}^{kx}\) and \(A\mathrm{e}^{kx}\)Multiply by \(k\): the derivative of \(5\mathrm{e}^{-2x}\) is \(-10\mathrm{e}^{-2x}\)
Grade C
Find the gradient at a given pointDifferentiate first, then substitute, giving exact answers such as \(2\mathrm{e}\)
Grade B
Find where the gradient has a given valueSolve an equation such as \(6\mathrm{e}^{2x} - 5 = 7\) using ln.
Grade A
Explain why exponential models suit many situationsFor \(y = A\mathrm{e}^{kx}\),\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = ky\): the rate of change is proportional to the amount present.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
The curve \(C\) has equation \(y = 3\mathrm{e}^{-2x} + 2\).
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).[1]
(b) Find the exact gradient of \(C\) at the point where \(x = 0\).[1]
(c) Find the exact coordinates of the point on \(C\) at which the gradient is \(-24\).[3]
Worked solution: The gradient of \(\mathrm{e}^{kx}\) is \(k\mathrm{e}^{kx}\), so \(\frac{\mathrm{d}y}{\mathrm{d}x} = -6\mathrm{e}^{-2x}\).
(b)Answer: \(-6\)
B1 for \(-6\) oe
Worked solution: When \(x = 0\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = -6\mathrm{e}^{0} = -6\).
(c)Answer: \(\left(-\ln 2, 14\right)\)
M1 for \(-6\mathrm{e}^{-2x} = -24\) leading to \(\mathrm{e}^{-2x} = 4\)
A1 for \(x = -\ln 2\)
A1 for \(y = 14\)
Worked solution: \(-6\mathrm{e}^{-2x} = -24 \Rightarrow \mathrm{e}^{-2x} = 4 \Rightarrow x = -\ln 2\). Then \(y = 3 \times 4 + 2 = 14\).
Question 2Medium7 marks
The curve \(C\) has equation \(y = \mathrm{e}^{x} + 2\). The point \(P\) on \(C\) has \(x\)-coordinate \(\ln 2\).
(a) Find the exact \(y\)-coordinate of \(P\) and the exact gradient of \(C\) at \(P\).[3]
(b) Find an equation of the tangent to \(C\) at \(P\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are exact constants.[2]
(c) The tangent meets the \(x\)-axis at the point \(Q\). Find the exact \(x\)-coordinate of \(Q\).[2]
Show the answer and mark scheme
(a)Answer: \(y = 4\), gradient \(2\)
B1 for \(y = 4\) (using \(\mathrm{e}^{\ln 2} = 2\))
M1 for \(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{x}\)
A1 for gradient \(2\)
Worked solution: \(\mathrm{e}^{\ln 2} = 2\), so \(y = 2 + 2 = 4\). \(\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{x}\), which at \(P\) is \(2\).
(b)Answer: \(y = 2x + 4 - 2\ln 2\)
M1 for \(y - 4 = 2(x - \ln 2)\)
A1 for \(y = 2x + 4 - 2\ln 2\)
Worked solution: \(y - 4 = 2(x - \ln 2) \Rightarrow y = 2x + 4 - 2\ln 2\)
(c)Answer: \(x = \ln 2 - 2\)
M1 for setting \(y = 0\) in their tangent
A1 for \(x = \ln 2 - 2\) oe
Worked solution: \(0 = 2(x - \ln 2) + 4 \Rightarrow x - \ln 2 = -2 \Rightarrow x = \ln 2 - 2\)
Question 3Hard7 marks
The curve \(C\) has equation \(y = \mathrm{e}^{2x}\). The point \(P\) on \(C\) has \(x\)-coordinate \(t\).
(a) Show that the tangent to \(C\) at \(P\) meets the \(x\)-axis at the point where \(x = t - \frac{1}{2}\).[3]
(b) Given that the tangent at \(P\) passes through the point \((1, 0)\), find the exact coordinates of \(P\).[2]
(c) The tangent at \(P\), the \(x\)-axis and the line \(x = \frac{3}{2}\) enclose a triangle. Find the exact area of this triangle.[2]
Show the answer and mark scheme
(a)Answer: Shown
M1 for the gradient \(2\mathrm{e}^{2t}\)
M1 for the tangent \(y - \mathrm{e}^{2t} = 2\mathrm{e}^{2t}(x - t)\) and setting \(y = 0\)
A1* for \(x = t - \frac{1}{2}\)
Worked solution: At \(P\), \(y = \mathrm{e}^{2t}\) and the gradient is \(2\mathrm{e}^{2t}\). Setting \(y = 0\) in the tangent: \(-\mathrm{e}^{2t} = 2\mathrm{e}^{2t}(x - t) \Rightarrow x - t = -\frac{1}{2}\), so \(x = t - \frac{1}{2}\).
Worked solution: The triangle has base from \(x = 1\) to \(x = \frac{3}{2}\), of length \(\frac{1}{2}\), and height \(\mathrm{e}^{3}\). Area \(= \frac{1}{2} \times \frac{1}{2} \times \mathrm{e}^{3} = \frac{\mathrm{e}^{3}}{4}\) (\(\approx 5.02\)).