calculus help please?

Question

its derivatives of logarithmic functions

how do you find y ' if x^y = y^x

Answer 1

you have to use implicit differentiation
x^y=y^x
ylnx=xlny
dy/dx(lnx) + (1/x)y= lny + (1/y)dy/dx(x)
y/x - lny = (x/y)dy/dx - (lnx)dy/dx
y/x - lny = ((x/y) -lnx)dy/dx
(y-xlny)/x = (x-ylnx)/y(dy/dx)
y(y-xlnx)/(x(x-ylnx) = dy/dx = y'

Answer 2

7

Answer 3

x^y = y^x when x = y
This gives a line of slope 1

The derivitive = 1

Answer 4

you want to first use the fact that log(a^b) = b*log a,
then take the derivative

Answer 5

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html

this web page looks pretty good.

after refreshing my memory, you basically just take the derivative of y in each term, so take d(x^y)/(dx)=d(y^x)/(dx)

Answer 6

Use logarithmic differentiation. First, take a logarithm of both sides.

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

Now take an implicit derivative:

y/x + y'*ln(x) = (x*y')/y + ln(y)

Solve for y'

y' * ln(x) - (x*y'/y) = ln(y) - y/x

y'(ln(x) - x/y) = ln(y) - y/x

y' = (ln(y) - y/x)/(ln(x) - x/y)

y' = (xy*ln(y) - y^2)/(xy*ln(x) - x^2)

y' = (y/x)*(x*ln(y)-y)/(y*ln(x) - x)

y'= (y/x) * (ln(y^x) - y)/(ln(x^y) - x)

That's about as clean as it's gonna get.