Differentiating e^x and ln x

The results

• d/dx (e^x) = e^x
• d/dx (e^kx) = ke^kx
• d/dx (ln x) = 1/x, for x > 0
• d/dx (ln kx) = 1/x as well, because ln kx = ln k + ln x

Inverse functions

e^{ln x} = x and ln(e^x) = x. The graph of y = ln x is the reflection of y = e^x in the line y = x.

Using them

These combine with the chain, product and quotient rules in P3. Always check what multiplies x inside the exponent.

Worked example

Find the gradient of y = 3e^2x at x = 0.

1. dy/dx = 3 x 2e^2x = 6e^2x
2. At x = 0: 6e⁰ = 6

Answer: The gradient is 6.

Check yourself

1. What is d/dx of e^5x?
   • e^5x
   • 5e^5x
   • 5xe^5x
   • e^5x/5
   Show answer

   5e^5x. Multiply by the coefficient of x: 5.

2. What is d/dx of ln(4x)?
   • 4/x
   • 1/(4x)
   • 4
   • 1/x
   Show answer

   1/x. ln 4x = ln 4 + ln x, and ln 4 is a constant, so the derivative is 1/x.

3. y = e^3x. Find dy/dx at x = 0.
   Show answer

   3. dy/dx = 3e^3x = 3 at x = 0.

4. y = 2 ln x. Find dy/dx at x = 4.
   Show answer

   0.5. dy/dx = 2/x = 0.5 at x = 4.

5. The graph of y = ln x is a reflection of y = e^x in which line?
   • y = x
   • The x-axis
   • The y-axis
   • y = -x
   Show answer

   y = x. Inverse functions reflect in y = x.

Practise with instant feedback

Sources

• Pearson Information Manual 2026/2027: International A level

Last updated 7 November 2026. Spotted a mistake? Email alevel@lifereadysociety.com.