Question

If \(x^m y^n=(x+y)^{m+n}\) then dy/dx=

Answer 1

\[(x^my^n)'\]
Use product rule here
Then there will be some power rule and chain rule involved for this

Answer 2

\[(fg)'=f'g+fg'\]

Answer 3

y^n=(x+y)^m+n/x^n

Answer 4

@myininaya does the queastion mean to just y

Answer 5

To find y'
I was going to differentiate both sides
Then isolate y'

Answer 6

okay

Answer 7

For the other side we will use chain rule
\[((h)^c)'=ch^{c-1}h'\]

Answer 8

I have solved it and i get the answer y/x but i am just searching for newer, better and shorter ways of solving it! I solved it by using partial differentiation.
What i did is:
\[\frac{dy}{dx}=-(\frac{y^n mx^{m-1}-(m+n)(x+y)^{m+n-1}}{x^mny^{n-1}-(m+n)(x+y)^{m+n-1}} )\]And then substituted values for \(x^m\) and \(y^n\) and did some calculations involving 4 steps (lengthy ones including lots of fractions) and i got y/x finally!

But i need to find a more easier way!

Ok, the options were also given:
a)-2x/y
b)x/y
c)y/x
d)x^y/2

I just need to be quick while solving these type of problems!
So, any ideas??

Answer 9

\[(x^my^n)'=(x^m)'y^n+x^m(y^n)' \text{ product rule}\]
\[((x+y)^{m+n})'=(m+n)(x+y)^{m+n-1}(x+y)' \text{ chain rule}\]

Answer 10

Now can you find the following:
\[(x^m)',(y^n)',(x+y)'?\]

Answer 11

@myininaya After doing what you say, i end up to what i have already posted. And i can reach there directly without using sum and product rule!

Answer 12

*chain

Answer 13

just take ln from both side

Answer 14

That gives me an answer 1.. @mukushla

Answer 15

can you do it for me?

Answer 16

just give me one sec ...maybe; this method will not work....

Answer 17

ok.

Answer 18

I even tried putting arbitrary values like for m and n and solving it. I had put m=1 and n=1 since m and n both are constants!
But that didn't help.

Answer 19

Hey the natural log thing works very well

Answer 20

@mukushla If there is + instead of - in the numerator in your above reply, it will give the answer!

Answer 21

\[x^my^n=(x+y)^{m+n}\]
Take natural log of both sides
\[m \ln(x)+n \ln(y)=(m+n) \ln(x+y)\]
Differentiate both sides
And then isolate y' (clear any resulting compound fractions you may have)
And then factor out y on top
And factor out x on bottom
Something cancels and there you have it! :)

Answer 22

Thanks @mukushla !! It works..
writing x instead of y earlier, made me get an answer 1.

I solved again and it worked!

Well, something still seems wrong in your above reply!

Answer 23

welcome my friend

Answer 24

Using natural log made it much easier and shorter! That was what i was searching for!

Answer 25

oh i solved it using log but i saw u already have it using that.
as u asked for diff method u can do this..
u can take a new variable k=x+y on left side and solve as

write k as x+y back now and use product rule on LHS to get the y' term and solve for y' ...