anonymous
  • anonymous
find dy/dx if e^(xy)=x+y+xy
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I just need help with the derivative of e^xy
SolomonZelman
  • SolomonZelman
what is the derivative of \(xy\) ?
anonymous
  • anonymous
y*dy/dx

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

SolomonZelman
  • SolomonZelman
use the product rule
anonymous
  • anonymous
on what term?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{dy}{dx}(f\cdot g)=f'g+fg' }\)
SolomonZelman
  • SolomonZelman
seen this before?
anonymous
  • anonymous
yup
SolomonZelman
  • SolomonZelman
Good
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{dy}{dx}(x+f(x))=xf'(x)+(x')f(x)=xf'(x)+f(x) }\)
SolomonZelman
  • SolomonZelman
am I right ?
anonymous
  • anonymous
I dont see how it equals the last part
SolomonZelman
  • SolomonZelman
The derivative of x is what?
anonymous
  • anonymous
oh I see yea
anonymous
  • anonymous
1
SolomonZelman
  • SolomonZelman
And now, do you see what I got before? \(\large\color{black}{ \displaystyle \frac{dy}{dx}(x+f(x))=xf'(x)+(x')f(x)=xf'(x)+f(x) }\)
SolomonZelman
  • SolomonZelman
agree or not?
anonymous
  • anonymous
agree
SolomonZelman
  • SolomonZelman
And same way, \(\large\color{black}{ \displaystyle \frac{dy}{dx}(x+y)=xy'+(x')y=xy'+y }\)
anonymous
  • anonymous
yup
SolomonZelman
  • SolomonZelman
Because y is really a function of x.
SolomonZelman
  • SolomonZelman
Ok, so what is the derivative of \(xy\) ?
anonymous
  • anonymous
x*y'+y
SolomonZelman
  • SolomonZelman
Ok, good
SolomonZelman
  • SolomonZelman
And you know that: \(\large\color{black}{ \displaystyle \frac{dy}{dx}e^{f(x)} =e^{f(x)}\cdot f'(x) }\) by the CHAIN RULE principal \(.....\) correct?
anonymous
  • anonymous
yup
SolomonZelman
  • SolomonZelman
Ok, so what do you get for: \(\large\color{black}{ \displaystyle \frac{dy}{dx}e^{xy} =? }\)
anonymous
  • anonymous
\[e ^{xy}*(x*\frac{ dy }{ dx }+y)\]
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
now we will differentiate the right side = x + y + xy
anonymous
  • anonymous
\[1+\frac{ dy }{ dx}+(x*\frac{ dy }{ dx}+y)\]
SolomonZelman
  • SolomonZelman
yes, very good
SolomonZelman
  • SolomonZelman
So, when you differentiate both sides, you get: \(\large\color{black}{ \displaystyle e^{xy}\left(x\frac{dy}{dx}+y\right)=1+\frac{dy}{dx}+x\frac{dy}{dx}+y }\)
anonymous
  • anonymous
yup
SolomonZelman
  • SolomonZelman
Isolate the \(\large\color{black}{ \displaystyle \frac{dy}{dx}, }\) (just using algebra)
anonymous
  • anonymous
ok
anonymous
  • anonymous
Did you also get this? \[\frac{ dy }{ dx }=\frac{ 1+ y}{ x e^{xy}-1-x }\]
SolomonZelman
  • SolomonZelman
I doubt that
anonymous
  • anonymous
oh yea I forgot something
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle e^{xy}\left(x\frac{dy}{dx}+y\right)=1+\frac{dy}{dx}+x\frac{dy}{dx}+y }\) \(\large\color{black}{ \displaystyle xe^{xy}\frac{dy}{dx}+ye^{xy}=1+\frac{dy}{dx}+x\frac{dy}{dx}+y }\) I will start you off
anonymous
  • anonymous
\[\frac{ dy }{ dx }=\frac{ 1+y+ye ^{xy} }{ xe ^{xy} -1-x}\]
anonymous
  • anonymous
woops the last term in numerator shouldbe negative
SolomonZelman
  • SolomonZelman
Yes, that is exactly correct \(\large\color{blue}{ \displaystyle \frac{ dy }{ dx }=\frac{ 1+y-ye ^{xy} }{ xe ^{xy} -1-x} }\)
SolomonZelman
  • SolomonZelman
Good job !
anonymous
  • anonymous
Awesome Thank You So Much! :)
SolomonZelman
  • SolomonZelman
Anytime !

Looking for something else?

Not the answer you are looking for? Search for more explanations.