# Find dy/dx

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For the equation: x^y=a^xy

logarithm of both sides will yield: y logx = xy log a

taking the derivative with respect to x of both sides and using multiplication rule, we get

y/x + dy/dx logx = loga[xdy/dx +y] = (x loga) dy/dx + y loga

or, dy/dx (logx - xloga) = y loga -y/x

or, **dy/dx = [y loga - y/x]/[logx - x loga]**

The given question is x^y = a^(x-y)

Taking the logarithm of both sides, y log x = (x-y) log a = x log a - y log a

Taking the derivative of both sides with respect to x and using the product rule,

dy/dx logx + y/x = log a -dy/dx log a

(remember derivative of a constant term is 0, so derivative of log a =0)

or, (log x + log a)dy/dx = log a - y/x

thus, **dy/dx = (log a - y/x)/(log x + log a)**

Its not minus on the power of rhs.it is a^(xy)