# `x^4 (x + y) = y^2 (3x - y)` Find `(dy/dx)` by implicit differentiation.

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### 2 Answers

**Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where 'n' = real number **

**2) 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^4)*(x + y) = (y^2)*(3x - y)

Differentiating both sides w.r.t 'x' we get

{4*(x^3)}*(x+y) + (x^4)*[1 + (dy/dx)] = {2y*(dy/dx)}*(3x-y) + (y^2)*{3 - (dy/dx)}

or, 4*(x^4) + 4y*(x^3) + (x^4) + (x^4)*(dy/dx) = 6xy*"(dy/dx) - 2(y^2)*(dy/dx) + 3*(y^2) - (y^2)*(dy/dx)

or, [5*(x^4) + 4y*(x^3) - 3*(y^2)] = (dy/dx)*[6xy - 3(y^2) - (x^4)]

or, dy/dx = [5*(x^4) + 4y*(x^3) - 3*(y^2)]/[6xy - 3(y^2) - (x^4)]

### User Comments

`x^4(x+y)=y^2(3x-y)`

`x^5+x^4y=3xy^2-y^3`

Differentiating with respect to x.We get

`5x^4+(4x^3y+x^4(dy/dx))=(3y^2+3x(2y)(dy/dx))-3y^2(dy/dx)`

`(5x^4+4x^3y-3y^2)=(6xy-3y^2-x^4)(dy/dx)`

`dy/dx=(5x^4+4x^3y-3y^2)/(6xy-3y^2-x^4)`

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