If y = a^x, where a > 0 and a ≠ 1, what is dy/dx?

If y = a^x, where a > 0 and a ≠ 1, what is dy/dx?

• A. a^x
• B. a^x ln(a)
• C. ln(a) / a^x
• D. x ln(a)

Answer: (B)

When differentiating an exponential with a constant base, use logarithms. Start with y = a^x and take natural logs: ln y = x ln a. Differentiate implicitly: (1/y) dy/dx = ln a. Multiply both sides by y to get dy/dx = y ln a, and substitute y = a^x to obtain dy/dx = a^x ln a. Since a > 0 and a ≠ 1, ln a is defined (and positive if a > 1, negative if 0 < a < 1), so the slope at any x is scaled by a^x. The expression a^x ln a is the correct rate of change. The other forms don’t come from this differentiation rule: they either miss the factor ln a, place it in the denominator, or replace the derivative with x times ln a, which would correspond to a different function.

When differentiating an exponential with a constant base, use logarithms. Start with y = a^x and take natural logs: ln y = x ln a. Differentiate implicitly: (1/y) dy/dx = ln a. Multiply both sides by y to get dy/dx = y ln a, and substitute y = a^x to obtain dy/dx = a^x ln a. Since a > 0 and a ≠ 1, ln a is defined (and positive if a > 1, negative if 0 < a < 1), so the slope at any x is scaled by a^x.

The expression a^x ln a is the correct rate of change. The other forms don’t come from this differentiation rule: they either miss the factor ln a, place it in the denominator, or replace the derivative with x times ln a, which would correspond to a different function.