Derivative of $f(x)^{g(x)}$

Question

Question: Find the derivative of

$$ f(x) = \left(\frac{1}{x}\right)^{\Large\frac{1}{x}} $$

Tip: convert $f(x)$ to $e^{g(x)}$.

How does one convert $f(x)$ to $e^{g(x)}$?

Thanks in advance

Answer 1

Consider the simple example $y = a^x$ where $a$ is a number. We have:

$$\begin{array}{ccc} y &=& a^x \, , \\ \ln y &=& x\ln a \, , \\ e^{\ln y} &=& e^{x\ln a} \, , \\ y &=& e^{x\ln a} \, . \end{array}$$

Thus, in your case $(1/x)^{1/x} \equiv e^{(1/x)\ln(1/x)} \equiv e^{-(\ln x)/x}.$ Now use the chain rule to to differentiate.

Answer 2

Hint: $$f(x)=e^{(1/x)\cdot\ln(1/x)}.$$ In your case, we can think of this as using the chain rule with $g(x)=e^{x\ln(x)}$ and $h(x)=1/x$ so that $f(x)=g\circ h(x)$. The chain rule implies that $f'(x)=g'(h(x))\cdot h'(x)$, thus \begin{align} f'(x)&=\left(\frac{1}{x}\ln(\frac{1}{x})\right)'e^{(1/x)\ln(1/x)}\\ \\ &=\left(-\frac{1}{x^2}\ln(1/x)-\frac{1}{x^2}\right)e^{(1/x)\ln(1/x)}\\ \\ &=\left(-\frac{\ln(1/x)+1}{x^2}\right)\cdot\left(\frac{1}{x}\right)^{1/x} \end{align}

Answer 3

In general, if $g(x)=\log f(x)$ then $f'(x) = g'(x) f(x)$. So you really only need to compute the derivative of $g(x)=\log (1/x)^{1/x} = \frac{-\log x}{x}$.

Answer 4

First, recall that for $\;h(x) = e^x, h^{-1}(x) = \ln x,\;$ so $\;h(h^{-1}(x)) = x$.

So, in this case, $\;f(x) = e^{\large\ln(f(x))}$

Then, to answer your question, with your function in hand: $$ \begin{align} f(x) &=\left(\frac{1}{x}\right)^{1/x} \\ \\ &=e^{\ln\left((1/x)^{\large 1/x}\right)} \\ \\ &= e^{\large(1/x)\cdot\ln(1/x)} \\ \\ &= e^{\large-\large(\ln x)/x} \end{align} $$ Now use the chain-rule to derivate.

Answer 5

Two ways come to mind without the rewrite... sorry I have answered the title rather than the actual question.

Let $y(x)=f(x)^{g(x)}$. Now take the natural logarithm of both sides:

$\ln(y(x))=\ln\left(f(x)^{g(x)}\right)=g(x)\ln(f(x))$.

This may only be done where $f(x)>0$ but for $y$ to be defined in the first place we need this anyway. Now differentiate both sides implicitly with respect to $x$ using power rule:

$\frac{1}{y(x)}\cdot\frac{dy}{dx}=g(x)\frac{1}{f(x)}\cdot f'(x)+g'(x)\cdot\ln(f(x))$.

Now multiply across by $y(x)$

$\frac{dy}{dx}=g(x)\cdot f(x)^{g(x)-1}\cdot f'(x)+f(x)^{g(x)}\cdot \ln(f(x))\cdot g'(x)$.

Alternate Solution

Let $y(f,g)=f^g$ where $f=f(x)$ and $g(x)$. Now use this Chain Rule

$\frac{d}{dt}f(x(t),y(t))=\frac{\partial f}{\partial x}\cdot\frac{dx}{dt}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dt}$

Thus we have

$\frac{dy}{dx}=(gf^{g-1})\cdot \frac{df}{dx}+(f^g\ln(f))\cdot \frac{dg}{dx}$

which is of course the same answer.