Differentiation of a function

Question

We need to find out the derivative $\frac{dy}{dx}$ of the following:

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

I know how to differentiate the function and on solving we get $\frac{dy}{dx}$=$\frac{y}{x}$. But we notice that $\frac{dy}{dx}$ is independent of values of $m$ and $n$. So what I did was again starting the problem from start but this time substituting any arbitrary values of $m$ and $n$, like $m$=$n$=$1$ (lets say). That is we get $xy$=$(x+y)^2$. Now opening the square bracket and solving we get

$x^2+y^2+xy=0$

So $\frac{dy}{dx}$=$\frac{-2x-y}{2y+x}$

But the answer should be $\frac{y}{x}$.Whats wrong in assuming arbitrary values of $m$ and $n$ if the answer dosen't depend upon values of $m$ and $n$.Please help.

How I got $\frac{dy}{dx}$=$\frac{y}{x}$?

$x+y$=$x^\frac{m}{m+n}$$y^\frac{m}{m+n}$

$ln(x+y)$=$\frac{m(ln(x))}{(m+n)}$+$\frac{n(ln(y))}{(m+n)}$

Differentiating we get

$\frac{1}{x+y}$.$(1+\frac{dy}{dx})$=$\frac{m}{(m+n)x}$+$\frac{n}{(m+n)y}\frac{dy}{dx}$

$\frac{dy}{dx}(\frac{1}{x+y}-\frac{n}{(m+n)y})$=$\frac{m}{(m+n)x}-\frac{1}{x+y}$

Just rearrange and you are done $\frac{dy}{dx}$=$\frac{y}{x}$

To me your result seems to be right. Why do you think that $dy/dx = y/x$? — MrYouMath, Mar 14 '17 at 14:18
I'm fairly certain the equation $x^2+y^2+xy=0$ only exists at $(0,0)$, so its probably a bad example. — Simply Beautiful Art, Mar 14 '17 at 14:37
Hm, it does appear you are right, though it takes quite a bit of work to get there. And upon solving the DE, you get $y=ax$ by the way. — Simply Beautiful Art, Mar 14 '17 at 15:10
@SimplyBeautifulArt No need to solve the DE by the way . See edits. — Piyush Raut, Mar 14 '17 at 15:15
@SimplyBeautifulArt Hey but please tell me whats wrong in assuming arbitrary values of m,n? Equation exists only at $(0,0)$ so its not working or there is some fault in my reasoning? — Piyush Raut, Mar 14 '17 at 15:18
You need to choose values of m,n such that the function actually exists lol. — Simply Beautiful Art, Mar 14 '17 at 15:25

Simply Beautiful Art · Answer 1 · 2017-03-14T14:31:41.443

I'm not quite sure how you got $\frac{dy}{dx}=\frac yx$. You should instead have

$$\frac d{dx}x^my^n=mx^{m-1}y^n+nx^my^{n-1}\frac{dy}{dx}$$

$$\frac d{dx}(x+y)^{m+n}=(m+n)(x+y)^{m+n-1}\left(1+\frac{dy}{dx}\right)$$

Put these two together and we get

$$mx^{m-1}y^n+nx^my^{n-1}\frac{dy}{dx}=(m+n)(x+y)^{m+n-1}\left(1+\frac{dy}{dx}\right)$$

Expand a bit:

$$mx^{m-1}y^n+nx^my^{n-1}\frac{dy}{dx}=(m+n)(x+y)^{m+n-1}+(m+n)(x+y)^{m+n-1}\frac{dy}{dx}$$

Group the $\frac{dy}{dx}$ terms:

$$\left(nx^my^{n-1}-(m+n)(x+y)^{m+n-1}\right)\frac{dy}{dx}=(m+n)(x+y)^{m+n-1}-mx^{m-1}y^n$$

Divide both sides:

$$\frac{dy}{dx}=\frac{(m+n)(x+y)^{m+n-1}-mx^{m-1}y^n}{nx^my^{n-1}-(m+n)(x+y)^{m+n-1}}$$

and I'm fairly certain this does not equal $\frac yx$. It does simplify a little though:

$$(x+y)^{m+n-1}=\frac{x^my^n}{x+y}$$

$$\frac{dy}{dx}=\frac{(m+n)\frac{x^my^n}{x+y}-mx^{m-1}y^n}{nx^my^{n-1}-(m+n)\frac{x^my^n}{x+y}}$$

Divide the numerator and denominator by $x^my^n$ to get

$$\frac{dy}{dx}=\frac{\frac{m+n}{x+y}-mx^{-1}}{ny^{-1}-\frac{m+n}{x+y}}$$

Now simplify the fractions:

$$\frac{dy}{dx}=\frac{(m+n)xy-my(x+y)}{nx(x+y)-(m+n)xy}$$

Expand and cancel like terms:

$$\frac{dy}{dx}=\frac{nxy-my^2}{nx^2-mxy}$$

Checking each result, it appears both come out to give the same conclusion of $y=ax$, though your answer was better in my opinion :-) — Simply Beautiful Art, Mar 14 '17 at 15:12

score 2 · Answer 2 · answered Mar 14 '17 at 14:43

2

Take $log$ on both sides and differentiate w.r.t $x$;

$\frac{m}{x} +\frac{n}{y}. y'=\frac{(m+n)}{(x+y)}. (1+y')$

$\implies y'(\frac{n}{y}-\frac{m+n}{x+y})=\frac{m+n}{x+y}-\frac{m}{x}$

$\implies y'=\frac{y}{x}$ on simplification

answered Mar 14 '17 at 14:43

Nitin Uniyal

7,946

Differentiation of a function

2 Answers2