Differentiation of exponential

[chaslltt]

Homework Statement

y = x^(18x^-4)

Homework Equations

chain rule
dy/dx a^x = a^x ln a

The Attempt at a Solution

first i used the second equation from above to get
x^(18/x^4)*ln(x)
then i use the chain rule to get
x^(18/x^4)*ln(x)*(-72x^-5)
the computer program i am using is saying this is wrong and the only thing i can think of that is wrong is the fact I may not be able to use the one formula since x is not a constant but if this is true then i have no idea what to do

 

[foxjwill]

Try taking the log of both sides and using implicit differentiation.

 

[chaslltt]

alright so then i should get
ln y = ln (x^(18/x^4))
1/y = 18/x^4 ln x
= y 18/x^4*1/x
= y 18/x^5
= x^(18/x^4)*18/x^5
?

 

[chaslltt]

well the computer told me that solution is wrong also so any other advice would be helpful

 

[Office_Shredder]

dy/dx a^x = a^x ln a

This is only true if a is a constant! Do you know how to differentiate x^x? Hint: Try writing it as e^f(x) for some function f(x). The same technique can be applied here.

It doesn't look like you took the derivative of the right hand side in your other post correctly

 

[chaslltt]

does this look right
ln y= 18/x^4 ln x
1/y= 18/x^5 + -72x^-5 ln x
= y 18/x^5 + -72x^-5 ln x
then just plug in y ?

 

[foxjwill]

does this look right
ln y= 18/x^4 ln x
1/y= 18/x^5 + -72x^-5 ln x
= y 18/x^5 + -72x^-5 ln x
then just plug in y ?

yep!

 

[HallsofIvy]

does this look right
ln y= 18/x^4 ln x
1/y= 18/x^5 + -72x^-5 ln x

You've forgotten the "dy/dx"!

= y 18/x^5 + -72x^-5 ln x

and what is this equal to?

then just plug in y ?

If $ln y= (18/x^4)ln(x)$
then [iitex](1/y)(dy/dx)= (18/x^5) -72x^{-5}ln x[/itex]
Now solve for dy/dx by multiplying both sides by y.
You might not be required to substitute for y. It depends on exactly what is wanted.