1. ## 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

2. ## 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

3. ## Re: Finding the derivative

with respect to y

4. ## 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'.

5. ## Re: Finding the derivative

x with respect to y. sorry, I meant y'

6. ## 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?

7. ## 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'

8. ## 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 ...
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}