Michael L.

asked • 09/28/17# how would i find the y = version of this expression?

i need to turn x

^{y}= y^{x}into a y = version so that i can take its derivative. how would i go about this?
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## 2 Answers By Expert Tutors

I am not sure that an explicit y = f(x) can be found for this problem, but the derivative can still be found for this problem.

ln(x

^{y}) = ln(y^{x})ylnx = xlny

Taking derivative of both sides with respect to x where y' = dy/dx

y'(lnx) + y(1/x) = lny + x(1/y)y'

(lnx - x/y)y' = lny - y/x

y' = (lny - y/x)/(lnx -x/y)

Arturo O. answered • 09/28/17

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Take natural logarithm of both sides.

ln(x

^{y}) = ln(y^{x})y ln(x) = x ln(y)

Now do implicit differentiation, using the chain rule.

y' ln(x) + y(1/x) = (1)ln(y) + x(1/y)y'

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

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

y'(x,y) = [ln(y) - y/x] / [ln(x) - x/y]

Now you need the particular point (x

_{1},y_{1}) where you want to evaluate y'.y'(x

_{1},y_{1}) = [ln(y_{1}) - y_{1}/x_{1}] / [ln(x_{1}) - x_{1}/y_{1}]I am not sure we can do much more without a pair of coordinates.

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Arturo O.

09/29/17