is there any horizantal tangent line in this function?

Question

y=x^√x

Answer 1

You can also use logarithmic differentiation.  This differentiation methods appreciates implicit differentiation.

 

Let  y=f(x)

 

y = x^√x

 

 

Log both sides of the equation and bring down any exponent as the coefficient of the log.

 

ln(y) = √(x)ln(x)

 

ln(y) = x^1/2ln(x)

 

 

Now, we can derive both sides of the equation.

 

y' / y = [(1/2)x^-1/2ln(x) + (√(x) / x)]

 

 

Multiply both sides of the equation by y.

 

y' = y * [(ln(x) / 2√x) + (√(x) / x)]

 

 

As a result when substituting, we get the derivative

 

f'(x) = x^√x * [(ln(x) / 2√x) + (√(x) / x)]

 

 

Now if you simplify, you should end up with the same result as Hassan's.

Answer 2

f(x)=x^√x

f(x)=e^√xlnx

f'(x)=(1/2√x + √x/x)x^√x

which is :

f'(x)=(lnx+2)/2√x  x^√x

f'(x)=0 means that lnx+2=0

therefore x=e^-2

then conclusion we have horizontal tangency at x=e^-2