Which argument correctly derives (d)/(dx)(e^kx)=ke^kx using the chain rule?

Set up the composite: y=e^u, u=kx Differentiate the outer function: (dy)/(du)=e^u Differentiate the inner function: (du)/(dx)=k Apply the chain rule: (dy)/(dx)=(dy)/(du)×(du)/(dx)=e^u× k Substitute back u=kx: (dy)/(dx)=ke^kx Restate what has been shown: Let u=kx; then (dy)/(dx)=e^u· k=ke^kx. Note it holds for all valid x: true for all x Link to the graph's behaviour: consistent with the graph Test with a numerical example: check a sample value Identify the matching option: Let u=kx; then (dy)/(dx)=e^u· k=ke^kx. Reject the first misconception: reject a common slip Reject the second misconception: reject a sign error Reject the third misconception: reject a rule mix-up Reject the fourth misconception: reject an over-generalisation Conclude: (d)/(dx)(e^kx)=ke^kx Answer: Let u=kx; then (dy)/(dx)=e^u· k=ke^kx.

The tangent to y=e^x at the point (2,e^2) meets the x-axis at A and the y-axis at B. Find the coordinates of A and of B.

Differentiate: (dy)/(dx)=e^x Gradient at x=2: m=e^2 Point-gradient form: y-e^2=e^2(x-2) Expand: y-e^2=e^2x-2e^2 Rearrange the tangent: y=e^2x-e^2 Find B on the y-axis: x=0: y=-e^2 Find A on the x-axis: y=0: 0=e^2x-e^2 Factor: 0=e^2(x-1) Solve: x=1 State A: A=(1, 0) State B: B=(0, -e^2) Collect the working so far: A(1, 0), B(0, -e^2) State the final answer clearly: A(1, 0), B(0, -e^2) Check the form of the answer: A(1, 0), B(0, -e^2) Summary: A(1,0), B(0,-e^2) Answer: A(1, 0), B(0, -e^2)

Hard A-Level Exponential functions Questions

Challenging, exam-style A-Level Exponential functions questions with worked solutions. Stretch yourself on the hardest tangent, exponential, coordinate geometry, intercept problems.

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A-Level34 questionsStep-by-step solutions

Question 1

6 markschallenging

Which graph represents

y=2-e^{-x}

Show worked solution

Worked solution

Find the y-intercept
$x=0:\ y=2-e^{0}=2-1=1$
Setting $x=0$ gives $y=1$ , so the curve passes through $(0,1)$ .
Find the asymptote
$x\to\infty:\ e^{-x}\to 0,\ y\to 2$
As $x$ grows, $e^{-x}\to0$ , so $y$ rises towards the asymptote $y=2$ .
Check behaviour for negative x
$x\to-\infty:\ e^{-x}\to\infty,\ y\to-\infty$
For very negative $x$ , $e^{-x}$ is huge, so $y$ dives downwards; the curve is increasing overall.
List the features to match
$\text{intercept, direction, asymptote}$
To choose the right graph we make a short checklist of the curve's key features: where it crosses the $y$ -axis, whether it rises or falls, and the position of its asymptote.
Check the y-intercept feature
$\text{correct }y\text{-intercept}$
The correct graph must cross the $y$ -axis at the height we calculated above. This rules out any option with the wrong starting point.
Check the direction feature
$\text{correct increasing/decreasing shape}$
The correct graph must rise or fall in the same way as our curve. A curve going the wrong way indicates a reflection, so it can be rejected.
Check the asymptote feature
$\text{correct horizontal asymptote}$
The correct graph must flatten out towards the same horizontal line we found. An option with a different asymptote has been shifted vertically and is wrong.
Reject the reflected graph
$\text{reject reflection}$
One distractor is the curve reflected in an axis, so it goes the wrong way. Because its direction disagrees with our analysis, we reject it.
Reject the vertically shifted graph
$\text{reject vertical shift}$
Another distractor has been moved up or down, giving the wrong asymptote and intercept. This does not match, so we reject it.
Reject the horizontally shifted graph
$\text{reject horizontal shift}$
A further distractor is shifted left or right, so its key point sits at the wrong $x$ -value. We reject it too.
Reject the non-exponential graph
$\text{reject wrong type}$
One option is not even the right type of curve (for example a straight line), so it cannot be an exponential graph and is rejected.
Confirm only one graph fits
$\text{one graph matches all features}$
Exactly one graph has the correct intercept, direction and asymptote all together, which pins down the answer.
Match to the correct graph
$\text{Increasing curve through }(0,1)\text{ with asymptote }y=2.$
We select the graph that matches every feature in our checklist.
Double-check with a sample point
$\text{verify one plotted point}$
As a final check we test one easy point on the chosen graph and confirm it agrees with our equation. It does, so the choice is secure.
Find the x-intercept
$y=0:\ e^{-x}=2\ \Rightarrow\ x=-\ln 2\approx -0.69$
The curve crosses the $x$ -axis where $e^{-x}=2$ , i.e. $x=-\ln2$ , just left of the origin.

Answer

\text{Increasing curve through }(0,1)\text{ with asymptote }y=2.

Question 2

6 markschallenging

Which argument correctly derives

\dfrac{d}{dx}\left(e^{kx}\right)=ke^{kx}

using the chain rule?

Show worked solution

Worked solution

Set up the composite
$y=e^{u},\ \ u=kx$
We treat $e^{kx}$ as a function of a function: the outer function is $e^{u}$ and the inner function is $u=kx$ .
Differentiate the outer function
$\frac{dy}{du}=e^{u}$
The derivative of $e^{u}$ with respect to $u$ is $e^{u}$ , since the exponential is its own derivative.
Differentiate the inner function
$\frac{du}{dx}=k$
Differentiating $u=kx$ with respect to $x$ gives the constant $k$ .
Apply the chain rule
$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=e^{u}\times k$
The chain rule multiplies the two derivatives together.
Substitute back u=kx
$\frac{dy}{dx}=ke^{kx}$
Replacing $u$ with $kx$ gives the required result $ke^{kx}$ .
Restate what has been shown
Let $u=kx$ ; then $\dfrac{dy}{dx}=e^{u}\cdot k=ke^{kx}$ .
We pause to state clearly the result our working has established. This is the statement we now match against the options.
Note it holds for all valid x
$\text{true for all } x$
The argument above did not rely on a special value of $x$ , so the conclusion holds generally. This generality is important for a correct proof or reasoning answer.
Link to the graph's behaviour
$\text{consistent with the graph}$
The conclusion is consistent with the shape of the exponential graph, which gives a helpful visual check on the algebra.
Test with a numerical example
$\text{check a sample value}$
Trying a simple number in the result confirms it behaves as claimed. A worked example like this reassures us the reasoning is right.
Identify the matching option
Let $u=kx$ ; then $\dfrac{dy}{dx}=e^{u}\cdot k=ke^{kx}$ .
Comparing our conclusion with the list of statements, exactly one option says precisely this. That is the answer.
Reject the first misconception
$\text{reject a common slip}$
One distractor comes from forgetting to bring the constant $k$ down when differentiating. Our working shows this is wrong.
Reject the second misconception
$\text{reject a sign error}$
Another distractor arises from a sign error or a reversed transformation. The steps above rule it out.
Reject the third misconception
$\text{reject a rule mix-up}$
A further distractor mixes up the exponential rule with the ordinary power rule. That does not apply here, so it is wrong.
Reject the fourth misconception
$\text{reject an over-generalisation}$
The last distractor over-generalises a special case that is not true in general. We reject it as well.
Conclude
$\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}$
This proves the gradient rule for $e^{kx}$ from the chain rule.

Answer

Let

u=kx

; then

\dfrac{dy}{dx}=e^{u}\cdot k=ke^{kx}

Question 3

6 markschallenging

Solve

e^{x}+6e^{-x}=5

, giving your answers in exact form.

Show worked solution

Worked solution

Clear the negative power
$e^{x}\cdot e^{x}+6e^{-x}\cdot e^{x}=5e^{x}$
Multiply every term by $e^{x}$ to remove the $e^{-x}$ . This works because $e^{-x}\cdot e^{x}=1$ .
Simplify each term
$e^{2x}+6=5e^{x}$
The first term becomes $e^{2x}$ , the middle $6\times1=6$ , and the right $5e^{x}$ .
Rearrange to standard form
$e^{2x}-5e^{x}+6=0$
Moving everything to one side gives a quadratic in $e^{x}$ .
Substitute u
$u=e^{x}\ \Rightarrow\ u^{2}-5u+6=0$
Letting $u=e^{x}$ makes the quadratic explicit.
Factorise
$\left(u-2\right)\left(u-3\right)=0$
Two numbers multiplying to 6 and adding to $-5$ are $-2$ and $-3$ .
Solve for u
$u=2\ \text{or}\ u=3$
Setting each bracket to zero.
First case
$e^{x}=2\ \Rightarrow\ x=\ln 2$
Taking logs of $e^{x}=2$ gives $x=\ln2$ .
Second case
$e^{x}=3\ \Rightarrow\ x=\ln 3$
Taking logs of $e^{x}=3$ gives $x=\ln3$ .
Check validity
$u=2,3>0\ \checkmark$
Both values are positive, so both give valid solutions for the exponential.
Collect the working so far
$x=\ln 2,\ x=\ln 3$
Before writing the answer, we gather the results of the previous lines to make sure nothing has been dropped. The key quantity we have found is shown above.
State the final answer clearly
$x=\ln 2,\ x=\ln 3$
We now write the answer on its own line. Setting it out clearly makes it easy to read back and easy for an examiner to mark.
Check the form of the answer
$x=\ln 2,\ x=\ln 3$
It is worth checking the answer has the expected form: exact values are left in terms of $e$ or $\ln$ , and coordinates are written as a pair. Ours matches, so the format is right.
Check the answer is reasonable
$x=\ln 2,\ x=\ln 3$
A quick sense-check against the size and sign expected from the question guards against slips. Everything is consistent here, so we can be confident.
Link back to the method
$x=\ln 2,\ x=\ln 3$
This result came directly from applying the exponential rules used above, so the logic is sound from start to finish.
State the solutions
$x=\ln 2\ \text{or}\ x=\ln 3$
So there are two exact solutions, $x=\ln2$ and $x=\ln3$ .

Answer

x=\ln 2,\ x=\ln 3

Question 4

6 markschallenging

An investment grows according to

V=2000e^{0.04t}

, where

V

is in pounds and

t

is in years. Find the exact time for the investment to double in value.

Show worked solution

Worked solution

State the doubling condition
$V=2\times 2000=4000$
Doubling means the value reaches twice the starting amount, so $V=4000$ .
Substitute into the model
$4000=2000e^{0.04t}$
We put $V=4000$ into the growth equation.
Divide by 2000
$2=e^{0.04t}$
Dividing both sides by the initial 2000 isolates the exponential and gives 2 on the left.
Take natural logs
$\ln 2=0.04t$
Taking $\ln$ brings the power down, since $\ln\left(e^{0.04t}\right)=0.04t$ .
Divide by 0.04
$t=\frac{\ln 2}{0.04}$
Dividing both sides by 0.04 isolates $t$ .
Simplify
$t=25\ln 2$
Since $\tfrac{1}{0.04}=25$ , this simplifies to $t=25\ln2$ .
Decimal check
$25\ln 2\approx 17.3$
Numerically about 17.3 years, which is a reasonable doubling time for 4% growth.
Note independence of starting value
$\text{doubling time does not depend on }2000$
Interestingly, the starting amount cancelled out, so any sum doubles in the same time under this model.
Collect the working so far
$25\ln 2$
Before writing the answer, we gather the results of the previous lines to make sure nothing has been dropped. The key quantity we have found is shown above.
State the final answer clearly
$25\ln 2$
We now write the answer on its own line. Setting it out clearly makes it easy to read back and easy for an examiner to mark.
Check the form of the answer
$25\ln 2$
It is worth checking the answer has the expected form: exact values are left in terms of $e$ or $\ln$ , and coordinates are written as a pair. Ours matches, so the format is right.
Check the answer is reasonable
$25\ln 2$
A quick sense-check against the size and sign expected from the question guards against slips. Everything is consistent here, so we can be confident.
Link back to the method
$25\ln 2$
This result came directly from applying the exponential rules used above, so the logic is sound from start to finish.
Record the result
$25\ln 2$
We box this off as the final result of the problem.
State the answer
$t=25\ln 2\ \text{years}$
So the investment doubles after exactly $25\ln2$ years.

Answer

25\ln 2

Question 5

6 markschallenging

The tangent to

y=e^{x}

at the point

\left(2,e^{2}\right)

meets the

x

-axis at

A

and the

y

-axis at

B

. Find the coordinates of

A

and of

B

Show worked solution

Worked solution

Differentiate
$\frac{dy}{dx}=e^{x}$
The gradient of the curve, and hence the tangent, is $e^{x}$ .
Gradient at x=2
$m=e^{2}$
At $x=2$ the gradient is $e^{2}$ .
Point-gradient form
$y-e^{2}=e^{2}\left(x-2\right)$
Using the contact point $\left(2,e^{2}\right)$ and gradient $e^{2}$ .
Expand
$y-e^{2}=e^{2}x-2e^{2}$
Multiplying $e^{2}$ through the bracket.
Rearrange the tangent
$y=e^{2}x-e^{2}$
Collecting constants: $-2e^{2}+e^{2}=-e^{2}$ .
Find B on the y-axis
$x=0:\ y=-e^{2}$
Setting $x=0$ gives the $y$ -intercept $y=-e^{2}$ , so $B=\left(0,-e^{2}\right)$ .
Find A on the x-axis
$y=0:\ 0=e^{2}x-e^{2}$
Setting $y=0$ to find where the tangent meets the $x$ -axis.
Factor
$0=e^{2}\left(x-1\right)$
Taking out the common factor $e^{2}$ .
Solve
$x=1$
Since $e^{2}\neq0$ , we get $x-1=0$ , so $x=1$ ; thus $A=(1,0)$ .
State A
$A=\left(1,\ 0\right)$
The tangent meets the $x$ -axis at $(1,0)$ .
State B
$B=\left(0,\ -e^{2}\right)$
And it meets the $y$ -axis at $\left(0,-e^{2}\right)$ .
Collect the working so far
$A\left(1,\ 0\right),\ B\left(0,\ -e^{2}\right)$
Before writing the answer, we gather the results of the previous lines to make sure nothing has been dropped. The key quantity we have found is shown above.
State the final answer clearly
$A\left(1,\ 0\right),\ B\left(0,\ -e^{2}\right)$
We now write the answer on its own line. Setting it out clearly makes it easy to read back and easy for an examiner to mark.
Check the form of the answer
$A\left(1,\ 0\right),\ B\left(0,\ -e^{2}\right)$
It is worth checking the answer has the expected form: exact values are left in terms of $e$ or $\ln$ , and coordinates are written as a pair. Ours matches, so the format is right.
Summary
$A\left(1,0\right),\ B\left(0,-e^{2}\right)$
So the two intercepts are $A=(1,0)$ and $B=\left(0,-e^{2}\right)$ .

Answer

A\left(1,\ 0\right),\ B\left(0,\ -e^{2}\right)

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