Question

how to find the derivative of e^xy = 2x+y

Answer 1

implicit differentiation for this one
use the chain rule and product rule

Answer 2

\[e^{xy}\times \left(y+xy'\right)=2+y'\] solve for \(y'\)

Answer 3

could u explain how to got y +xy'

Answer 4

yes

Answer 5

you are taking the derivative of \(e^{xy}\) with respect to \(x\)
the derivative of \(e^{whatever}\) is \(e^{whatever}\times (whatever)'\)so your real job is to take the derivative of \(xy\) with respect to \(x\)
for this you need the product rule

Answer 6

we are thinking of \(y\) as a function of \(x\) so it is like taking the derivative of \[xf(x)\] by the product rule you get
\[f(x)+xf'(x)\] but we are writing \(y\) instead of \(f(x)\) and \(y'\) instead of \(f'(x)\)
write instead
\[y+xy'\]

Answer 7

ok. Would you mind helping with another problem ?

Answer 8

sure

Answer 9

i'm trying to find the derivative of e^x/ 1-e^2x , i did it, but I get a different answer than the book

Answer 10

\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=e^x, f'(x)=e^x, g(x)=1-e^{2x}, g'(x)=-2e^{2x}\]

Answer 11

before any simplification you get
\[\frac{(1-e^{2x})e^x-e^x\times (-2e^{2x})}{(1-e^{2x})^2}\]

Answer 12

can simplify this a bit for sure

Answer 13

yes that is what I have. Can you simplify it, bc when I do it i get a wrong answer

Answer 14

\[\frac{(1-e^{2x})e^x-e^x\times (-2e^{2x})}{(1-e^{2x})^2}\]
\[\frac{(1-e^{2x})e^x+e^x\times (2e^{2x})}{(1-e^{2x})^2}\]

Answer 15

\[\frac{e^x(1-e^{2x}+2e^{2x})}{(1-e^{2x})^2}\]

Answer 16

wait how did you get -2e^2x before simplifying

Answer 17

\[\frac{e^x(1+e^{2x})}{(1-e^{2x})^2}\]

Answer 18

wait, I know

Answer 19

I just missed a negative when I was taking the derivative of the bottom

Answer 20

ahh i see

Answer 21

Sorry to ask, but could you help me with another question