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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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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