For those what are good at math

[twitchee2]

Im having trouble figuring what the answer is this this equation

d/dx x^(1-x)

After I worked it out, i came up with the solution of: (-lnx+1/x^-1)x^(1-x)

Can anyone verify that for me please. Thanks.

 

[soydios]

chain rule is your friend!

outside function: exponential (x^u)
remember that the derivative of exponential function "a^b" is "(a^b)ln(a)"
thus part of the answer will be: "(x^(1-x))ln(x)"

inside function: exponent (1-x)
derivative of constant "1" is zero, derivative of "-x" is -1
thus the other part of the answer will be: "(-1)"

final answer: "(x^(1-x))ln(x)(-1)"

 

[BigJ]

y = x^(1-x)
ln (y) = (1-x) * ln(x)


Use Chain Rule:
1/y * y' = -1 * ln(x) + 1/x * (1-x)
y' = (-1 * ln(x) + 1/x * (1-x)) * y

Remember y = x^(1-x)

So y':
y' = (-1 * ln(x) + 1/x * (1-x)) * x^(1-x)

You can clean it up a bit if you want, but that answer should suffice.

 

[twitchee2]

Here is what i did, im kind of lost from what you did.

x^(1-x) = e^1-xlnx then d/dx e^u = u'e^u.

set u = 1-xlnx

the derivative of 1-xlnx come out to be -lnx+1/x-1 correct?

then plugging everything back in i get : (-lnx+1/x-1)e^1-xlnx which simplifies to (-lnx+1/x-1)x^(1-x) <-- my final answer. does this semm correct?

 

[Rob9874]

Answer: 42

 

[BigJ]

Originally posted by: BigJ
y = x^(1-x)
ln (y) = (1-x) * ln(x)


Use Chain Rule:
1/y * y' = -1 * ln(x) + 1/x * (1-x)
y' = (-1 * ln(x) + 1/x * (1-x)) * y

Remember y = x^(1-x)

So y':
y' = (-1 * ln(x) + 1/x * (1-x)) * x^(1-x)

You can clean it up a bit if you want, but that answer should suffice.

BTW, this answer is correct. Did it by hand and checked it in my Ti-89. Ti-89 Simplifies it to (-x *ln(x) - x + 1) * x^(-x) through some nifty algebra.

http://www.analyzemath.com/calculus/Differentiation/first_derivative.html

Thats a link on how to do it by hand.

 

[twitchee2]

awesome thanks man.

 

[dogooder]

Originally posted by: twitchee2
Here is what i did, im kind of lost from what you did.

x^(1-x) = e^((1-x)lnx) then d/dx e^u = u'e^u.

set u = (1-x)lnx

the derivative of (1-x)lnx come out to be -lnx+1/x-1 correct?

then plugging everything back in i get : (-lnx+1/x-1)e^((1-x)lnx) which simplifies to (-lnx+1/x-1)x^(1-x) <-- my final answer. does this semm correct?

I added parentheses, but it's correct.