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

2 Answers
May 20, 2018

dy/dx = (-y - e^y/sqrt(x-y)) /(1 + x + 2e^y sqrt(x-y) - e^y/(sqrt(x-y)))dydx=yeyxy1+x+2eyxyeyxy

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)))

May 20, 2018

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)))