$$\frac{d}{dx} y^x$$

How would you find the derivative with respect to $x$ of $y^x$ assuming that $y$ is a function of $x$? I know you will have to use the chain rule somehow, and I know that the derivative of $a^x = a^x \cdot\log (a)$.

Taking the derivative of the outer function, $y^x$, gives $y^x\log(x)\frac{\text{dy}}{\text{dx}}$, and the derivative of the inner function, $y$ gives $\frac{dy}{dx}$.

Putting this together gives

$$\frac{\text{d}}{\text{dx}} y^x=\frac{\text{dy}}{\text{dx}}x^y \log(x) \cdot \frac{dy}{dx}$$

Is this even right? How should I proceed further?

  • 1
    $\begingroup$ Did you mean $\dfrac{d}{dx}y^x$ or derivative of $\dfrac{dy}{dx}y^x$? $\endgroup$ – Tunk-Fey May 22 '14 at 3:42
  • $\begingroup$ @Tunk-Fey The former $\endgroup$ – 1110101001 May 22 '14 at 3:42
  • $\begingroup$ @DonAntonio Even if y is a function of x? Must not the chain rule be used in that case? $\endgroup$ – 1110101001 May 22 '14 at 3:43
  • 1
    $\begingroup$ You can't use that formula, because it's meant for when a is a constant, not a variable. See my answer. $\endgroup$ – Deepak May 22 '14 at 3:45
  • $\begingroup$ Then the answer is $y^x\left(\ln y+\dfrac{x}{y}\dfrac{dy}{dx}\right)$ by chain rule. $\endgroup$ – Tunk-Fey May 22 '14 at 3:51

$$\frac{d}{dx} y^x\\ = \frac{d}{dx}e^{x\ln y} = e^{x\ln y}.(\ln y + \frac{x}{y}.\frac{dy}{dx})\\ = y^x.(\ln y + \frac{x}{y}.\frac{dy}{dx})$$

(basically converting to an exponential function, then applying chain rule and product rule).

  • $\begingroup$ Is it not possible to directly take the derivative without converting to the exponential form? $\endgroup$ – 1110101001 May 22 '14 at 3:45
  • $\begingroup$ Not directly, no. You can recast it as an equation, viz. $y^x = z$ take logs of both sides, and differentiate both sides wrt x implicitly. But it comes to the same thing, really, just more unwieldy. BTW, what I did is the "standard trick" for functions raised to other functions. $\endgroup$ – Deepak May 22 '14 at 3:48

We want to find $\frac{dz}{dx}$, where $z=y^x$.

Taking the natural logarithm of both sides, we get $$\ln z=x\ln y.$$ Now differentiate. We get $$\frac{1}{z}\frac{dz}{dx}=\frac{x}{y}\frac{dy}{dx}+\ln y.$$ Since $z=y^x$, we get $$\frac{dz}{dx}=y^x\left( \frac{x}{y}\frac{dy}{dx}+\ln y \right).$$

Remark: The procedure we used is called logarithmic differentiation. It is a technique that is really not needed here, but is useful, for example, if we want the derivative of an ugly product like $(\cos x)(1+3x)^{12}(1+x^2)^{17}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.