`x e^y = x - y` Find `(dy/dx)` by implicit differentiation.
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hkj1385
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Note:- 1) If y = e^x ; then dy/dx = e^x
2) If y = x^n ; then dy/dx = n*x^(n-1) ; where 'n' = real number
3) If y = u*v ; where both u & v are functions of 'x' ; then
dy/dx = u*(dv/dx) + v*(du/dx)
Now, the given function is :-
x*(e^y) = x - y
Differentiating both sides w.r.t 'x' we get
x*(e^y)*(dy/dx) + (e^y) = 1 + (dy/dx)
or, [x*(e^y) - 1]*(dy/dx) = [1 - (e^y)]
or, dy/dx = [1 - (e^y)]/[x*(e^y) - 1]
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balajia | Student
`xe^y=x-y`
Differentiating both sides with respect to x.
`e^y+xe^y(dy/dx)=1-dy/dx`
`(1+xe^y)(dy/dx)=1-e^y`
`dy/dx=(1-e^y)/(1+xe^y) `
Student Answers