Question

If `sqrt(1+y)+ysqrt(1+x)=0`, then `dy/dx=`?

Answer 1

See Below.

Explanation:

We have,

`color(white)(xxx)sqrt(1 + y) + ysqrt(1 + x) = 0`

`rArr ysqrt(1 + x) = -sqrt(1 + y)`

`rArr y^2(1 + x) = 1 + y` [Squaring both sides]

`rArr y^2(1 + x) - y - 1 = 0`

So, `D = b^2 - 4ac = 1 + 4 * (1 + x) = 4(1 + x)`

And `y = (-b+- sqrt(D))/(2a) = (1 +-sqrt(4(1 + x)))/(2(1 + x)) = (1 + 2sqrt(1 + x))/(2(1 + x)), (1 - 2sqrt(1 +x))/(2(1 + x))`

So,

`dy/dx = d/dx(1 +-sqrt(4(1 + x)))/(2(1 + x))= -((x + 1)^(3/2) + x^2 + 2x + 1)/(2(x + 1)^(7/2)),-((x + 1)^(3/2) - x^2 - 2x - 1)/(2(x + 1)^(7/2))`

[Used The Derivative Calculator Online. Didn't have too much time. Anyone can edit and add the explanation, if you please.]

Hope this helps.