Question

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

Answer 1

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

Explanation:

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)`

Answer 2

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!