# Find dy/dx by implicit differentiation. sqrt(5x+y) = 6+x^2y^2

### 2 Answers | Add Yours

sqrt(5x+y) = 6 + x^2 *y^2

Square both sides:

==> (5x + y ) = 36 + 12x^2 y^2 + x^4 y^4

==> x^4y^4 + 12x^2 y^2 - 5x - y + 36 = 0

==> (xy)^4 + 12(xy)^2 - 5x -y + 36 = 0

Now let us differentiate:

==> 4(xy)^3 (xy)' + 24(xy)(xy)' - 5 - y' + 0 = 0

==> 4(xy)^3 (y+xy') + 24(xy)(y+xy') -5 - y' = 0

==> 4x^3 y^4 + 4x^4 y^3 y' + 24xy^2 + 24x^ 2y y' - 5 - y' = 0

Now keep terms with y' on one side:

==> 4x^4 y^3 y' + 24x^2 y y' - y' = 5- 24xy^2 - 4x^3 y^4

Now factor y':

==> y'(4x^4 y^3 + 24x^2 y -1) = (5-24xy^2 -4x^3 y^4)

**==> y' = (5-24xy^2-4x^3y^4)/(4x^4 y^3 + 24x^2 y - 1)**

To differentiate sqr(5x+y) = 6+x^2y^2).

We differentiate both sides of the given equation with respect to x and then solve for dy/dx or y'

Now we diffrentiate :

{sqrt(5x+y)}' = (6+x^2y^2)'.

(1/2)(5x+y)^(1/2 -1) *(5x+y)' = 0+ 2x*y^2 +x^2*2yy'

(1/2)(5x+y)^(-1/2) *(5+y') = 2xy^2 + 2x^2yy'

(1/2)(5x+y)^(-1/2)*y' -2x^2yy' = 2xy^2 -(1/2)(5x+y)^(-1/2)

y' {(1/2)(5x+y)^(-1/2) -2x^2y} = {2xy^2 -(1/2)(5x+y)^(-1/2)}

y' = {2xy^2 -(1/2)(5x+y)^(-1/2)}/{(1/2)(5x+y)^(-1/2) -2x^2y}

Multiply the numerator and denominator on the right side by 2*(5x+y)^(1/2).

y' = {2xy^2*sqrt(5x+y) -1}/{1-2x^2*sqrt(5x+y)}