Question

differentiate y=x^y with respect to x

Answer 1

The equation in your question actually defines two different functions:

One, which could be expressed explicitly. is the trivial one, namely y=x, and along that one the answer to your question is clearly y’=1.

The second one is an implicit function y=y(x) which we we are unable to describe explicitly in terms of elementary functions. However, by a well known theorem in calculus of functions of several variables, known as the Implicit Function Theorem, once an implicit function y=f(x) is given by an equation of the form F(x,y)=0, where F(x,y) is a given function with continuous partial derivatives with respect to both variables x and y, and its partial derivative with respect to y is non vanishing at a point (a,b) which satisfies F(a,b)=0, i.e. f(a)=b, then the implicit function y=f(x) is differentiable at x=a with derivative given by:

y’(a)=-(dF(a,b)/dx)/(dF(a,b)/dy)(a,b) (where dF(a,b)/dx and dF(a,b)/dy standing for the 2 partial derivatives of F(x,y) at (a,b) )

In the case we are talking about, it would be easier to write the equation x^y=y^x as F(x,y)=y*log(x)-x*log(y)=0. Then we can easily apply the Implicit Function Theorem to that equation, and to any point (a,b) satisfying F(a,b)=0 (that is a^b=b^a ), provided that a and b are unequal (e.g. a=2, b=4) , and since

dF/dx(a,b)=b/a-log(b)=(b-a*log(b))/a=(b-b*log(a))/a=b*(1-log(a)/a,

dF/dy(a,b)=log(a)-a/b=(b*log(a)-a)/b=(a*log(b)-a)/b=a*(log(b)-1)/b

we deduce:

y’(a)=-((b/a)^2)*(1-log(a))/(log(b)-1)

For example, for a=2, b=4 one gets

y’(2)=-4*(1-log(2))/(log(4)-1)=-3.17739889912418….

Answer 2

$dy \div dx \: \: = 2x$

dy/dx

dy/dx FORMULA

y = nX^ n-1

ANSWER

2x

HOPE IT HELPS YOU