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

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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...

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neela | Student

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)}

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