Question

solve the following

Answer 1

\[e^{xy} = e^{4x} - e^{5y}\]

Answer 2

Take ln each term.
This cancels out the e^(...)

Answer 3

So, taking ln of all terms leads
xy=4x-5y
Now solve for whatever term.

Answer 4

\[\frac{ d }{ dx }e^{4x} = e^{u}*du/dx = 4e^{4x}\]

\[\frac{ d }{ dx }e^{5y} = 5e^{5y}*\frac{ dy }{ dx }\]

\[4e^{4x}+5e^{5y}*\frac{ dy }{ dx }\]

Answer 5

whaa

Answer 6

this is implicit differentiation @sh3lsh sorry should have said so earlier

Answer 7

The other side

\[\frac{ d }{ dx } e^{xy} = \frac{ d }{ dx }e^{x'y}+\frac{ d }{ dx }e ^{xy' }\]

Answer 8

\[e^{xy} + \frac{ dy }{ dx }e^{xy} = 4e^{4x} + 5e^{5y}\frac{ dy }{ dx }\]

Answer 9

\[e^{xy}(1 + \frac{ dy }{ dx }) = 4e^{4x} + 5e^{5y}\frac{ dy }{ dx }\]

Answer 10

Is there something i'm missing here?

Answer 11

Hi

Answer 12

hi :)

Answer 13

Is this solved?

Answer 14

no got stuck, don't know how to solve this. it was implicit differentiation

Answer 15

With respect to what variable?

Answer 16

\[e^{xy} = e^{4x} - e^{5y}\]

Answer 17

well the problem says treat y as a function of x

Answer 18

I'm guessing we differentiate with respect to \(x\).

Answer 19

Let's start with the left hand side.

Answer 20

yep, I tried doing that then got suck

Answer 21

ok

Answer 22

Well, let me try to make it easier then.

Answer 23

Let \(u=xy\), this means \(e^{xy}=e^u\).
\[
\frac{d}{dx}e^u = \frac{d}{du}e^u\frac{du}{dx}
\]Does this make sense?

Answer 24

Yes, i'm following

Answer 25

For the first part:\[
\frac{d}{du}e^u=e^u=e^{xy}
\]For the second part:\[
\frac{du}{dx} = \frac{d}{dx}xy=\frac{dx}{dx}y+x\frac{dy}{dx}=y + x\frac{dy}{dx}
\]

Answer 26

Before I thought it was product rule

Answer 27

Which part? We have to start using the chain rule.

Answer 28

For some reason I can't see what you just wrote let me refresh my page

This was what I originally did.

\[e^{xy}(1 + \frac{ dy }{ dx }) = 4e^{4x} + 5e^{5y}\frac{ dy }{ dx }\]


For the other side this was what I had.

\[\frac{ d }{ dx }e^{4x} = e^{u}*du/dx = 4e^{4x}\]

\[\frac{ d }{ dx }e^{5y} = 5e^{5y}*\frac{ dy }{ dx }\]

\[4e^{4x}+5e^{5y}*\frac{ dy }{ dx }\]

Answer 29

You didn't do the left side correctly

Answer 30

This is what I thought: I thought that \[e^{xy}\] meant that we had to do product rule

Answer 31

oh wait... no

Answer 32

hmmm....I ended up with 0

Answer 33

you see @wio @jdoe0001 showed me the chain rule earlier. I dont know why I was thinking product rule

Answer 34

\[
e^{xy}
\]Is not a product, it's an exponent if anything.

Answer 35

hmm wait a sec... wrote it off the beaten path =)

Answer 36

so it's kind of the same like you said before right?

\[e^{u}, du/dx \]

Answer 37

Yes

Answer 38

\[e^{xy} = e^{xy}*\frac{ du }{ dx }\]

\[u = xy, = e^{xy}*y + x \frac{ dy }{ dx }\]

Answer 39

That is what I wrote earlier, but you need parenthesis.

Answer 40

for some reason I can't see it

Answer 41

Maybe you need to refresh the page. It shows just find for me

Answer 42

\[ e^{xy}(y+(\frac{ dy }{ dx })) = 4e^{4x} + 5e^{5y}\frac{ dy }{ dx }\]

Answer 43

I have to keep on refreshing my page

Answer 44

hmmm

Answer 45

You forgot the \(x\) before the derivative.

Answer 46

anyhow @greatlife44 \(\Large \cfrac{d}{dx}[e^{{\color{brown}{ f(x)}}}]\implies {\color{brown}{ f'(x)}}e^{{\color{brown}{ f(x)}}}\)

Answer 47

\[e^{xy}(y+x(\frac{ dy }{ dx })) = 4e^{4x} + 5e^{5y}\frac{ dy }{ dx }\]

Answer 48

yeap

Answer 49

ok so I had to do a-lot of rearranging lol

Answer 50

ehehh

Answer 51

\[\frac{ ye^{xy}-4e^{4x} }{ 5e^{5y}-xe^{xy} } = \frac{ dy }{ dx }\]

Answer 52

I wish there was a faster way for me to type equations on here so I could show you guys all the steps I did.

Answer 53

well, looks good \(\bf e^{xy} = e^{4x} - e^{5y}
\\ \quad \\
\left( y+x\cfrac{dy}{dx}\right)e^{xy}=4e^{4x}-5\cfrac{dy}{dx}e^{5y}
\\ \quad \\
ye^{xy}+xe^{xy}\cfrac{dy}{dx}=4e^{4x}-5\cfrac{dy}{dx}e^{5y}
\\ \quad \\
\cfrac{dy}{dx}\left( xe^{xy}+5e^{5y} \right)=4e^{4x}-ye^{xy}
\\ \quad \\
\cfrac{dy}{dx}=\cfrac{4e^{4x}-ye^{xy}}{xe^{xy}+5e^{5y}}\)

Answer 54

I got a 0 before, because I was doing some LN stuff lol

Answer 55

Thanks guys, it's unfortunate i can only give one medal :(

Answer 56

HERE IT IS

\[\frac{ 4e^{4x} }{ e^{xy} } - \frac{ 5e^{5y} }{ e^{xy} }\frac{ dy }{ dx } = y +\frac{ dy }{ dx }\]

\[\frac{ 4e^{4x} }{ e^{xy} } - y = \frac{ dy }{ dx }(x)+\frac{ 5e^{5y} }{ e^{xy} }\frac{ dy }{ dx }\]

\[\frac{ 4e^{4x} }{ e^{xy} } - y = \frac{ dy }{ dx }((x)+\frac{ 5e^{5y} }{ e^{xy} })\]

\[\frac{ \frac{ 4e^{4x} }{ e^{xy} }-y }{ \frac{ 5e^{xy} }{ e^{xy} }+x } =\frac{ dy }{ dx }\]

\[\frac{ e^{xy}*\frac{ 4e^{4x} }{ e^{xy} }-y*e^{xy} }{ e^{xy}*\frac{ 5e^{xy} }{ e^{xy} }+x*e^{xy} } =\frac{ dy }{ dx }\]

\[\frac{ 4*e^{4x}-y*e^{xy} }{ 5*e^{xy}+x*e^{xy} } = \frac{ dy }{ dx }\]

OMG YES!