What is the derivative of #y= (sqrt x) ^x#?

Question

Scarlett Allison · Accepted Answer

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

We can write the function as:

#(sqrtx)^x= (x^(1/2))^x= x^(x/2)=e^((xlnx)/2)#

Now:

#d/(dx) (sqrtx)^x = d/(dx)e^((xlnx)/2) = e^((xlnx)/2)*d/(dx)(xlnx)/2=1/2e^((xlnx)/2) (x*1/x+lnx)=1/2(sqrtx)^x(1+lnx)#

Emilia Aguilar · Answer

Take the natural logarithm (written #ln#) of both sides.

#lny = ln(sqrt(x))^x#

To get rid of the exponent, use the power rule of logarithms that states that #lna^n = nlna#

#lny = xlnsqrt(x)#

Differentiate the left hand side using implicit differentiation and the right hand using the chain and product rules. Before using the product rule, we must use the chain rule.

let #y = lnu# and #u = sqrt(x)#. Then #dy/(du) = 1/u# and #(du)/dx = 1/(2sqrt(x))#

Call #f(x) = ln(sqrt(x))#.

#f'(x) = dy/(du) xx (du)/dx#

#f'(x) = 1/u xx 1/(2sqrt(x))#

#f'(x) = 1/sqrt(x) xx 1/(2sqrt(x))#

#f'(x) = 1/(2x)#

Now to the rest of the function:

#1/y(dy/dx) = 1(lnsqrt(x)) + x(1/(2x))#

#1/y(dy/dx) = lnsqrt(x) + 1/2#

#dy/dx= (sqrt(x))^x(lnsqrt(x) + 1/2)#

Hopefully this helps!

Christopher Agresta · Answer

undefined To find the derivative of $ y = (\sqrt{x})^x $, you can use logarithmic differentiation. The steps are as follows:

1. Take the natural logarithm of both sides of the equation.
2. Apply the logarithmic properties to simplify the expression.
3. Differentiate both sides with respect to $ x $.
4. Solve for $ \frac{{dy}}{{dx}} $.

Starting with $ y = (\sqrt{x})^x $:

$$ \ln(y) = \ln\left((\sqrt{x})^x\right) $$

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

Now, differentiate both sides with respect to $ x $:

$$ \frac{1}{y} \frac{dy}{dx} = \ln(\sqrt{x}) + \frac{x}{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} $$

$$ \frac{1}{y} \frac{dy}{dx} = \ln(\sqrt{x}) + \frac{x}{2x} $$

$$ \frac{1}{y} \frac{dy}{dx} = \ln(\sqrt{x}) + \frac{1}{2} $$

Now, solve for $ \frac{dy}{dx} $:

$$ \frac{dy}{dx} = y \left( \ln(\sqrt{x}) + \frac{1}{2} \right) $$

$$ \frac{dy}{dx} = (\sqrt{x})^x \left( \ln(\sqrt{x}) + \frac{1}{2} \right) $$

Thus, the derivative of $ y = (\sqrt{x})^x $ is $ \frac{dy}{dx} = (\sqrt{x})^x \left( \ln(\sqrt{x}) + \frac{1}{2} \right) $.