Find dy/dx by implicit differientiation. `(x+y)^3+x^3+y^3=0`i have no idea how to take the derivative of y

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lemjay | High School Teacher | (Level 3) Senior Educator

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`(x+y)^3 + x^3+y^3=0`

Using implicit differentaition, there is no need to solve for y in order to get dy/dx. Instead, we directly take the derivateive of both sides with respect to x.

`d/(dx) [ (x+y)^3 + x^3 + y^3]=d/(x) (0)`

For the right side, note that the derivative of a constant is equal to the constant itselft ( `d/(du) (c) = c` ). 

`d/(dx)[ (x+y)^3 + x^3 + y^3]=0`

`d/(dx) (x+y)^3 + d/(dx)(x^3) + d/(dx) (y^3) = 0`

For the left side, apply the power formula of derivatives which is `d/(du) (u^n) = n*u^(n-1)* u' ` .

`3(x+y)^2* d/(dx) (x+y) + 3x^2 *d/(dx)(x) + 3y^2d/(dx)(y) = 0`

`3(x+y)^2 (d/(dx)(x)+d/(dx)(y)) + 3x^2 d/(dx)(x)+ 3y^2d/(dx)(y)=0`

`3(x+y)^2 (1 + (dy)/(dx)) + 3x^2+3y^2(dy)/(dx)=0`

At the left side, 3 is a common factor. So to simplify the equation, divide both sides by 3.

`(3(x+y)^2 (1 + (dy)/(dx)) + 3x^2+3y^2(dy)/(dx))/3=0/3`

`(x+y)^2 (1 + (dy)/(dx)) + x^2+y^2(dy)/(dx)=0`

`(x+y)^2 + (x+y)^2(dy)/(dx) + x^2+y^2(dy)/(dx)=0`

Then, isolate `(dy)/(dx)` . So, move `(x+y)^2`  and` x^2`  to the right side.

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

Factor out `(dy)/(dx)` .

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

Then, move `((x+y)^2 + y^2)` to the right.

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

Expand `(x+y)^2` .

`(dy)/(dx)= (-(x^2+2xy+y^2)-x^2)/(x^2+2xy+y^2+y^2)`

`(dy)/(dx)=(-x^2-2xy-y^2-x^2)/(x^2+2xy+y^2+y^2)`

`(dy)/(dx)=(-2x^2-2xy-y^2)/(x^2+2xy+2y^2)`

`(dy)/(dx)= - (2x^2+2xy+y^2)/(x^2+2xy+2y^2)`

Hence, `(dy)/(dx)= - (2x^2+2xy+y^2)/(x^2+2xy+2y^2)` .

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