I need help with yet another implicit differentiation e^x/y=x-y I've tried ((1/y)(e^(x/y))((y-xy')/(y^2)))=1-y' (e^(x/y))-1=(-y'y^2)/(y-xy') What am I doing wrong? I don't see a way
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Originally Posted by ppark I need help with yet another implicit differentiation e^x/y=x-y I've tried ((1/y)(e^(x/y))((y-xy')/(y^2)))=1-y' (e^(x/y))-1=(-y'y^2)/(y-xy') What am I doing wrong? I don't see a way I presume this is Can you take it from there? -Dan
Originally Posted by topsquark I presume this is Can you take it from there? -Dan Does the 1/y on e^(x/y) not need to come down when taking the derivative of e EX: d/dx e^(x/y)= 1/y e^(x/y)
Originally Posted by ppark Does the 1/y on e^(x/y) not need to come down when taking the derivative of e EX: d/dx e^(x/y)= 1/y e^(x/y) It's there. Try it using the chain rule and put u = x/y. Then <--- This is correct version of the term you wanted to "bring down." So we get as before. -Dan
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