What is the derivative of #x^(3/x)#?

Question

Isabella Ackerman · Accepted Answer

#"d"/("d"x) x^(3/x) = (ln(3/x)-3)*x^(3/x)# Write #x^(3/x) = exp(-xln(x/3))#. Then, by the chain rule, #"d"/("d"x) x^(3/x) = exp(-xln(x/3))*"d"/("d"x)(-xln(x/3))#, #"d"/("d"x) x^(3/x) = exp(-xln(x/3))*(-ln(x/3)-x*1/(x/3))#, #"d"/("d"x) x^(3/x) = exp(-xln(x/3))*(ln(3/x)-3)#, #"d"/("d"x) x^(3/x) = (ln(3/x)-3)*x^(3/x)#.

Aurora Alvarado · Answer

undefined To find the derivative of $x^{3/x}$, you can use the logarithmic differentiation method.

Let $y = x^{3/x}$.

Take the natural logarithm of both sides:

$$\ln(y) = \ln(x^{3/x})$$

Apply the power rule of logarithms:

$$\ln(y) = \frac{3}{x} \ln(x)$$

Differentiate implicitly with respect to $x$:

$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left(\frac{3}{x} \ln(x)\right)$$

Using the product rule and the chain rule:

$$\frac{1}{y} \frac{dy}{dx} = \frac{3}{x^2} \ln(x) - \frac{3}{x^2}$$

Now, substitute back $y = x^{3/x}$:

$$\frac{1}{x^{3/x}} \frac{dy}{dx} = \frac{3}{x^2} \ln(x) - \frac{3}{x^2}$$

Multiply both sides by $x^{3/x}$:

$$\frac{dy}{dx} = x^{3/x} \left(\frac{3}{x^2} \ln(x) - \frac{3}{x^2}\right)$$

Simplify:

$$\frac{dy}{dx} = 3x^{1 - \frac{3}{x}} (\ln(x) - 1)$$

So, the derivative of $x^{3/x}$ is $3x^{1 - \frac{3}{x}} (\ln(x) - 1)$.