Passage: Let (e^y )^x-y=0 , where y is a function of x whose domain is (0,10] . Question: What is dy dx equal to, given that y=y₀ when x=1 ?

Step-by-step solution

Given the equation (e^y)^x - y = 0 , we can rewrite it as: e^ xy = y Taking the natural logarithm on both sides: xy = y Differentiating both sides with respect to x using the product rule and chain rule: y + x dy dx = 1 y dy dx Rearranging the terms to solve for dy dx : y = ( 1 y - x ) dy dx y = ( 1 - xy y ) dy dx dy dx = y^2 1 - xy We are given that at x = 1 , y = y₀ . Substituting these values into the derivative: dy dx = y₀^2 1 - y₀ From the original equation e^ xy = y , substituting x = 1 and y = y₀ gives: e^ y