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Exercise 14

Avg difficulty 2.5

Implicit differentiation enables us a different perspective from which to see why the rule ddx[ax]=axln⁡(a)\frac{d}{dx} [a^x] = a^x \ln(a) holds, if we assume that ddx[ln⁡(x)]=1x\frac{d}{dx}[\ln(x)] = \frac{1}{x}. This exercise leads you through the key steps to do so.

  1. Let y=axy = a^x. Rewrite this equation using the natural logarithm function to write xx in terms of yy (and the constant aa).

  2. Differentiate both sides of the equation you found in (a) with respect to xx, keeping in mind that yy is implicitly a function of xx.

  3. Solve the equation you found in (b) for dydx\frac{dy}{dx}, and then use the definition of yy to write dydx\frac{dy}{dx} solely in terms of xx. What have you found?

Boelkins, Active Calculus, §2.7 Derivatives of functions given implicitly, Exercise 14

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.