Know that the gradient of e^(kx) is equal to k e^(kx) and hence understand…

The idea

Differentiating y = e^(kx) gives dy/dx = k e^(kx): the function reproduces itself multiplied by the constant k. This follows from the chain rule, since the derivative of e^u with respect to u is e^u and u = kx has derivative k. The result means the rate of change is proportional to the current value, dy/dx = ky, which is exactly the relationship modelled by growth and decay processes such as populations, compound interest, radioactive decay and cooling. When k > 0 the quantity grows; when k < 0 it decays. You can verify the rule numerically by comparing gradients from a table of values, and you use it to find gradients, tangents and stationary behaviour of exponential curves.

What earns the marks

• States the derivative of e^(kx) as k e^(kx) and identifies k as a constant multiplier arising from the chain rule.
• Shows the chain-rule reasoning: differentiate the outer exponential, then multiply by the derivative of the inner function kx.
• Links the result to the differential equation dy/dx = ky, explaining that rate of change is proportional to the quantity present.
• Interprets the sign of k: positive k gives growth, negative k gives decay, and connects this to real contexts such as population or radioactive decay.
• Applies the rule to find gradients, equations of tangents or normals, and to locate and classify stationary points where they exist.
• Uses the rule with a composite inner function, for example differentiating e^(3x + 1) by recognising k = 3.

Common mistakes and how to put them right

• Forgetting to multiply by k and writing the derivative as e^(kx).

  Put it right by applying the chain rule fully so the factor k appears.

• Differentiating the exponent as though it were a power of x, for example treating e^(kx) like x^n.

  Put it right by keeping the exponential form unchanged.

• Losing the negative sign when k is negative, which reverses growth and decay.

  Put it right by substituting k carefully, for example k = −0.05 gives −0.05e^(−0.05x).

• Assuming every exponential curve has a stationary point.

  Put it right by noting k e^(kx) is never zero for finite x, so no turning points occur.

Exam tips

• Write down k explicitly before differentiating, especially when the exponent is a compound expression such as 4x − 7.
• If a question describes a rate proportional to the amount present, translate it immediately into dy/dx = ky and identify k from the context.
• Check the sign of your derivative against the expected growth or decay described in the question.
• When finding a tangent, evaluate both the function and its derivative at the given x-value before forming the line equation.