Find the deivative of Y= e^bx^2 and the one of Ln(x+y) =e ^x/y?

Question

Y=e^(bx^2)

and Ln(x+y) = e^(x/y)

Answer 1

y = e^(bx^2)
dy/dx = (2bx)(e^(bx^2))

Ln(x+y) = e^(x/y)
d/dx Ln(x+y) = d/dx (e^(x/y))
1/(x+y) + 1/(x+y) dy/dx = (1/y)e^(x/y) - (x/y^2)e^(x/y) dy/dx
1/(x+y) dy/dx + (x/y^2)e^(x/y) dy/dx = (1/y)e^(x/y) - 1/(x+y)
[1/(x+y) + (x/y^2)e^(x/y)] dy/dx = (1/y)e^(x/y) - 1/(x+y)
dy/dx = [(1/y)e^(x/y) - 1/(x+y)] / [1/(x+y) + (x/y^2)e^(x/y)]

Answer 2

Y= e^bx^2

d/dxY=2bxe^bx^2 (using chain rule + remember that d/dx e^x =e^x or d/dx e^ax=ae^ax

Ln(x+y) =e ^x/y?
if u want d/dx then d/dx ln(x+y)=(1/(x+y) )(dy/dx) ...(chain rule)

and d/dx e ^x/y?=(1/y + (-x/y^2)dy/dx)e ^x/y?....(chain rule)

just equate the 2 and solve for dy/dx .

Answer 3

In the first question you haven't indicated with either brackets or multiplication sign so it's difficult to see what you are actually asking

Answer 4

y= e^bx^2

write down

y'=(2bx)*e^(bx^2)

Ln(x+y) =e ^(x/y)
let partial wrt to x =fx'
let partial wrt to y =fy'

ln(x+y) -e^(x/y)=0

fx'= 1/(x+y)-e^(x/y)/y
fy'= 1/(x+y)+x*e^(x/y)/y^2

dy/dx= -fx'/fy'
{fx'+fy'(dy/dx)=0,standard
relationship}
therefore,
dy/dx={e^(x/y)/y-1/(x+y)}
/{x*e^(x/y)/y^2+1/(x+y)}

i hope that this helps