If \(x^m y^n=(x+y)^{m+n}\) then dy/dx=
\[(x^my^n)'\] Use product rule here Then there will be some power rule and chain rule involved for this
\[(fg)'=f'g+fg'\]
y^n=(x+y)^m+n/x^n
@myininaya does the queastion mean to just y
To find y' I was going to differentiate both sides Then isolate y'
okay
For the other side we will use chain rule \[((h)^c)'=ch^{c-1}h'\]
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??
\[(x^my^n)'=(x^m)'y^n+x^m(y^n)' \text{ product rule}\] \[((x+y)^{m+n})'=(m+n)(x+y)^{m+n-1}(x+y)' \text{ chain rule}\]
Now can you find the following: \[(x^m)',(y^n)',(x+y)'?\]
@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!
*chain
just take ln from both side
That gives me an answer 1.. @mukushla
can you do it for me?
just give me one sec ...maybe; this method will not work....
ok.
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.
Hey the natural log thing works very well
@mukushla If there is + instead of - in the numerator in your above reply, it will give the 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! :)
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!
welcome my friend
Using natural log made it much easier and shorter! That was what i was searching for!
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' ...
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