# Finding the derivative

• Nov 20th 2012, 01:39 PM
NettieL
Finding the derivative
Hey there,

I am having trouble understanding how to work out this question, which is x(x+y)-e^y =y^2

I have been given the working out, of which the first line is (x+y)+x(1+y^1)-e^y *y^1=2yy^1
Really not to sure what rules they used to get from a to b, so if anyone could share, that would be most appreciated :)

Nettie.L
• Nov 20th 2012, 01:44 PM
Re: Finding the derivative
Are you differentiating with respect to y or with respect to x?
Overall, in order to solve the problem you have to have knowledge of what is known as Implicit Differentiation
• Nov 20th 2012, 01:56 PM
NettieL
Re: Finding the derivative
with respect to y
• Nov 20th 2012, 02:05 PM
HallsofIvy
Re: Finding the derivative
Do you mean you want to differentiate x with respect to y or y with respect to x? The first is what you said but what you give would be correct if by "y^1" you mean the derivative of y with respect to x, normally written as y'. Solve for y'.
• Nov 20th 2012, 02:06 PM
NettieL
Re: Finding the derivative
x with respect to y. sorry, I meant y' :)
• Nov 20th 2012, 02:24 PM
skeeter
Re: Finding the derivative
the derivative of y w/r to x is y' or dy/dx

the derivative of x w/r to y is x' or dx/dy

one more time ... which?
• Nov 20th 2012, 02:37 PM
NettieL
Re: Finding the derivative
well, in the worked answer they have d/dy (so I'm guessing that means dx/dy). What I really want to know is how they get from that first question, to the (x+y)+x(1+y')-e^y *y'=2yy'
• Nov 20th 2012, 03:50 PM
skeeter
Re: Finding the derivative
the original is taking the derivative w/r to x , so y is treated as a function of x

$\frac{d}{dx} \left[x(x+y)-e^y =y^2\right]$

product rule with the first term, chain rule w/ the rest ...

$(1)(x+y) + x(1 + y') - e^y \cdot y' = 2y \cdot y'$

the goal now is to algebraically solve for y', the derivative of y w/r to x ...