Question

find dy/dx if e^(xy)=x+y+xy

Answer 1

I just need help with the derivative of e^xy

Answer 2

what is the derivative of \(xy\) ?

Answer 3

y*dy/dx

Answer 4

use the product rule

Answer 5

on what term?

Answer 6

\(\large\color{black}{ \displaystyle \frac{dy}{dx}(f\cdot g)=f'g+fg' }\)

Answer 7

seen this before?

Answer 8

yup

Answer 9

Good

Answer 10

\(\large\color{black}{ \displaystyle \frac{dy}{dx}(x+f(x))=xf'(x)+(x')f(x)=xf'(x)+f(x) }\)

Answer 11

am I right ?

Answer 12

I dont see how it equals the last part

Answer 13

The derivative of x is what?

Answer 14

oh I see yea

Answer 15

1

Answer 16

And now, do you see what I got before?

\(\large\color{black}{ \displaystyle \frac{dy}{dx}(x+f(x))=xf'(x)+(x')f(x)=xf'(x)+f(x) }\)

Answer 17

agree or not?

Answer 18

agree

Answer 19

And same way,

\(\large\color{black}{ \displaystyle \frac{dy}{dx}(x+y)=xy'+(x')y=xy'+y }\)

Answer 20

yup

Answer 21

Because y is really a function of x.

Answer 22

Ok, so what is the derivative of \(xy\) ?

Answer 23

x*y'+y

Answer 24

Ok, good

Answer 25

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?

Answer 26

yup

Answer 27

Ok, so what do you get for:

\(\large\color{black}{ \displaystyle \frac{dy}{dx}e^{xy} =? }\)

Answer 28

\[e ^{xy}*(x*\frac{ dy }{ dx }+y)\]

Answer 29

yes

Answer 30

now we will differentiate the right side
= x + y + xy

Answer 31

\[1+\frac{ dy }{ dx}+(x*\frac{ dy }{ dx}+y)\]

Answer 32

yes, very good

Answer 33

So, when you differentiate both sides, you get:

\(\large\color{black}{ \displaystyle e^{xy}\left(x\frac{dy}{dx}+y\right)=1+\frac{dy}{dx}+x\frac{dy}{dx}+y }\)

Answer 34

yup

Answer 35

Isolate the \(\large\color{black}{ \displaystyle \frac{dy}{dx}, }\) (just using algebra)

Answer 36

ok

Answer 37

Did you also get this?
\[\frac{ dy }{ dx }=\frac{ 1+ y}{ x e^{xy}-1-x }\]

Answer 38

I doubt that

Answer 39

oh yea I forgot something

Answer 40

\(\large\color{black}{ \displaystyle e^{xy}\left(x\frac{dy}{dx}+y\right)=1+\frac{dy}{dx}+x\frac{dy}{dx}+y }\)

\(\large\color{black}{ \displaystyle xe^{xy}\frac{dy}{dx}+ye^{xy}=1+\frac{dy}{dx}+x\frac{dy}{dx}+y }\)

I will start you off

Answer 41

\[\frac{ dy }{ dx }=\frac{ 1+y+ye ^{xy} }{ xe ^{xy} -1-x}\]

Answer 42

woops the last term in numerator shouldbe negative

Answer 43

Yes, that is exactly correct

\(\large\color{blue}{ \displaystyle \frac{ dy }{ dx }=\frac{ 1+y-ye ^{xy} }{ xe ^{xy} -1-x} }\)

Answer 44

Good job !

Answer 45

Awesome Thank You So Much! :)

Answer 46

Anytime !