Solving a function

Question

Let y= (√x)^x

(a) Explain (in words) why the power rule cannot be applied the way the function
is written.

(b)Explain (in words) why the exponential derivative rule (i.e. d/dx (a^x) = a^x ln(a)) cannot be applied the way the function is written.

(c)Explain (in words) how we have to solve this problem? Your answer should state why your method will work.

(d) Find y'

Answer 1

The answer to the first two is about like - "I run out of fingers to count on".  Basically, the problem is too hairy and contains components that don't play nice using those rules.  You need to explain what those components are.

 

 

Okay, I'm going to skip straight to the third one.  Let's rephrase "the square root of X" to be X raised to the 1/2 power.    

 

Y = (sqrt(x))^X  = ( X^(1/2) )^X  = X^(1/2)*X   = X ^X/2

 

Now, that's almost something we can work with.  

 

Now, remember that X = e ^logX, So

 

Y =  X^X/2 =  (e ^logX )^X/2 = e ^X/2*logX 

 

In that form, you can use the exponential derivative rule and then the product rule to get Y'.

 

When Y = e^f(x)  Then Y' = f'(x)*e^f(x)  (the exponential derivative rule)

 

All you have to do is find the derivative of X/2*log(X)

 

Remember that  when H(X) = F(X)G(X)  then H'(X) = F(X)G'(X) + F'(X)G(X) (the product rule)

 

Can you solve it now?

 

 

 

 

 

Answer 2

Lets rewrite the function to get rid of the square-root symbol.

 

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

 

a)

The power rule cannot be applied the way the function is written because you have a function within a function. This means that you have to evaluate the square-root of x first.  Then you evaluate that result by raising it to the x power.   This is know as the chain rule. We have a linear function and an exponential function linked together. Suppose x = 4.

 

y = (√4)⁴

 

y = (2)⁴  ---->  we evaluate the square-root of 4 first.

 

y = 16 ------->  we raise the previous result to the 4th power.

 

 

b)

The exponential rule cannot be applied the way the function is applied because the base number a is some constant.  The function we have has a base of x.  x is a variable which can be any value.

 

c)

The method we should use is logarithmic differentiation along with implicit differentiation in which we take the natural log of both sides of equation because logarithms is the technique used to find the value of the exponent.

 

d)

 

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

 

ln y = (1/2)x lnx

 

(ln y)' = [(1/2)x lnx]'

 

(1/y)y' = [(1/2)lnx + (1/2)x (1/x)]

 

       y' = y[(1/2)lnx + (1/2)]

 

 

Hopefully, this helps answer your question.  These types of derivatives are really not that straight forward.  But once you know how the functions relate to one another, you can devise a method to deal with them.