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

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Answer 1 · 2016-12-21T14:18:30

#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)#

Answer 2 · 2016-12-21T14:19:11

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!