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

Question

myininaya · Answer

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

myininaya · Answer

$$(fg)'=f'g+fg'$$

anonymous · Answer

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

anonymous · Answer

@myininaya does the queastion mean to just y

myininaya · Answer

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

anonymous · Answer

okay

myininaya · Answer

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

ujjwal · Answer

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??

myininaya · Answer

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

myininaya · Answer

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

ujjwal · Answer

@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!

ujjwal · Answer

*chain

anonymous · Answer

just take ln from both side

ujjwal · Answer

That gives me an answer 1.. @mukushla

ujjwal · Answer

can you do it for me?

anonymous · Answer

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

ujjwal · Answer

ok.

ujjwal · Answer

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.

myininaya · Answer

Hey the natural log thing works very well

ujjwal · Answer

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

myininaya · Answer

$$x^my^n=(x+y)^{m+n}$$
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! :)

ujjwal · Answer

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!

anonymous · Answer

welcome my friend

ujjwal · Answer

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

anonymous · Answer

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 
|dw:1342472594684:dw|
write k as x+y back now and use product rule on LHS to get the y' term and solve for y' ...