Question

How do you implicitly differentiate `-y=xy+2e^ysqrt(x-y)`?

Answer 1

`dy/dx = (-y - e^y/sqrt(x-y)) /(1 + x + 2e^y sqrt(x-y) - e^y/(sqrt(x-y)))`

Explanation:

start by differentiating both sides w.r.t x

`therefore d/dx (-y) = d/dx (xy + 2e^(y)sqrt(x-y))`

lets look at this term by term,
RHS,
the derivative of xy with respect to x using the product rule comes as `d/dx (x) * y + x*d/dx (y)` which is equal to `y + x dy/dx`

we need to use the chain rule and the product rule for
differentiating `2e^y sqrt(x-y)`

= `d/dx (2e^y)*sqrt(x-y) + 2e^y*d/dx (sqrt(x-y))`

the derivative of `2e^y` w.r.t x is `2e^y dy/dx`

and the derivative of `sqrt(x-y) = (x-y)^(1/2)` using the chain rule comes as `1/(2sqrt(x-y))*(1-dy/dx)`

therefore the derivative of the second term is
`2e^y dy/dx*sqrt(x-y)+(2e^y)/(2sqrt(x-y))*(1-dy/dx)`

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

therefore, the entire equation becomes

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

multiply both the sides by -1 to make the left hand side positive,

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

Now shift all the dy/dx terms to the left side and factorise

#dy/dx + x dy/dx + 2e^y sqrt(x-y)dy/dx - e^y/(sqrt(x-y)) dy/dx = -y - e^y/(sqrt(x-y))#

`dy/dx (1 + x + 2e^y sqrt(x-y) - e^y/(sqrt(x-y))) = -y - e^y/(sqrt(x-y)`

therefore,
`dy/dx = (-y - e^y/sqrt(x-y)) /(1 + x + 2e^y sqrt(x-y) - e^y/(sqrt(x-y)))`

Answer 2

The answer is `=-(y+e^y*1/(sqrt(x-y)))/(1+x+2e^ysqrt(x-y)-e^y/(sqrt(x-y)))`

Explanation:

Let,

`f(x,y)=y+xy+2e^ysqrt(x-y)`

And

`dy/dx=-((delf) /(delx))/((delf) /(dely))`

`(delf) /(delx)=y+2e^y*1/(2sqrt(x-y))`

`(delf) /(dely)=1+x+2e^ysqrt(x-y)-e^y/(sqrt(x-y))`

Therefore,

`dy/dx=-(y+e^y*1/(sqrt(x-y)))/(1+x+2e^ysqrt(x-y)-e^y/(sqrt(x-y)))`