How do you find the derivative of #y=x^(ln(x))#?

Question

Cameron Alexander · Accepted Answer

#d/dx(x^lnx) = 2lnx x^(lnx-1)#

Write the function as:

#y= x^(lnx) = (e^lnx)^lnx = e^(ln^2x)#

then using the chain rule:

#d/dx(x^lnx) = d/dx (e^(ln^2x)) = e^(ln^2x) d/dx (ln^2x) = (2e^(ln^2x)lnx)/x#

and simplifying:

#d/dx(x^lnx) = 2lnx x^(lnx-1)#

Scarlett Allison · Answer

undefined To find the derivative of $ y = x^{\ln(x)} $, you can use the logarithmic differentiation technique.

Start by taking the natural logarithm of both sides of the equation to simplify it:

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

Then, apply the logarithmic property that allows you to bring the exponent down:

$$ \ln(y) = (\ln(x)) \cdot (\ln(x)) $$

Now, differentiate both sides of the equation with respect to $ x $:

$$ \frac{d}{dx}(\ln(y)) = \frac{d}{dx}(\ln(x) \cdot \ln(x)) $$

Use the chain rule and product rule on the right side:

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

Substitute $ y = x^{\ln(x)} $ back into the equation:

$$ \frac{1}{x^{\ln(x)}} \cdot \frac{dy}{dx} = \frac{1}{x} \cdot \ln(x) + \frac{1}{x} \cdot \ln(x) $$

Finally, solve for $ \frac{dy}{dx} $ to find the derivative:

$$ \frac{dy}{dx} = x^{\ln(x)} \cdot \left(\frac{1}{x} \cdot \ln(x) + \frac{1}{x} \cdot \ln(x)\right) $$

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

So, the derivative of $ y = x^{\ln(x)} $ is $ \frac{dy}{dx} = x^{\ln(x)} \cdot \frac{2\ln(x)}{x} $.