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Question Number 84059 by niroj last updated on 09/Mar/20

Find (dy/dx) of x^m x^n =(x+y)^(m+n) .

$$ \\ $$$$\:\boldsymbol{\mathrm{Find}}\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:\boldsymbol{\mathrm{of}}\:\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}} \boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} \:=\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)^{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} . \\ $$

Commented by jagoll last updated on 09/Mar/20

x^(m+n) = (x+y)^(m+n) (m+n) x^(m+n−1) = (m+n) (x+y)^(m+n−1) +( m+n)(x+y)^(m+n−1) y ′ x^(m+n−1) = (x+y)^(m+n−1) (1+y ′) 1+y′ = (x^(m+n−1) /((x+y)^(m+n−1) )) y′ = ((x^(m+n−1) −(x+y)^(m+n−1) )/((x+y)^(m+n−1) )) y ′ = (((x^(m+n) /x)−(((x+y)^(m+n) )/(x+y)))/(((x+y)^(m+n) )/(x+y))) = (((x^(m+n) /x)−(x^(m+n) /(x+y)))/(x^(m+n) /(x+y))) = (((1/x)−(1/(x+y)))/(1/(x+y))) = (x+y) ((y/(x(x+y)))) = (y/x)

$${x}^{{m}+{n}} \:=\:\left({x}+{y}\right)^{{m}+{n}} \\ $$$$\left({m}+{n}\right)\:{x}^{{m}+{n}−\mathrm{1}} \:=\:\left({m}+{n}\right)\:\left({x}+{y}\right)^{{m}+{n}−\mathrm{1}} +\left(\:{m}+{n}\right)\left({x}+{y}\right)^{{m}+{n}−\mathrm{1}} \:{y}\:' \\ $$$${x}^{{m}+{n}−\mathrm{1}} \:=\:\left({x}+{y}\right)^{{m}+{n}−\mathrm{1}} \:\left(\mathrm{1}+{y}\:'\right)\: \\ $$$$\mathrm{1}+\mathrm{y}'\:=\:\frac{\mathrm{x}^{\mathrm{m}+\mathrm{n}−\mathrm{1}} }{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{m}+\mathrm{n}−\mathrm{1}} } \\ $$$$\mathrm{y}'\:=\:\frac{\mathrm{x}^{\mathrm{m}+\mathrm{n}−\mathrm{1}} −\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{m}+\mathrm{n}−\mathrm{1}} }{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{m}+\mathrm{n}−\mathrm{1}} } \\ $$$$\mathrm{y}\:'\:=\:\frac{\frac{\mathrm{x}^{\mathrm{m}+\mathrm{n}} }{\mathrm{x}}−\frac{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{m}+\mathrm{n}} }{\mathrm{x}+\mathrm{y}}}{\frac{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{m}+\mathrm{n}} }{\mathrm{x}+\mathrm{y}}}\:=\:\frac{\frac{\mathrm{x}^{\mathrm{m}+\mathrm{n}} }{\mathrm{x}}−\frac{\mathrm{x}^{\mathrm{m}+\mathrm{n}} }{\mathrm{x}+\mathrm{y}}}{\frac{\mathrm{x}^{\mathrm{m}+\mathrm{n}} }{\mathrm{x}+\mathrm{y}}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}}{\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}}\:=\:\left(\mathrm{x}+\mathrm{y}\right)\:\left(\frac{\mathrm{y}}{\mathrm{x}\left(\mathrm{x}+\mathrm{y}\right)}\right)\:=\:\frac{\mathrm{y}}{\mathrm{x}} \\ $$

Commented by MJS last updated on 09/Mar/20

let m+n=q x^q =(x+y)^q ⇒ y=0 if we stay within R

$$\mathrm{let}\:{m}+{n}={q} \\ $$$${x}^{{q}} =\left({x}+{y}\right)^{{q}} \\ $$$$\Rightarrow\:{y}=\mathrm{0}\:\mathrm{if}\:\mathrm{we}\:\mathrm{stay}\:\mathrm{within}\:\mathbb{R} \\ $$

Answered by som(math1967) last updated on 09/Mar/20

(dy/dx)=(y/x)

$$\frac{{dy}}{{dx}}=\frac{{y}}{{x}} \\ $$

Commented by niroj last updated on 09/Mar/20

Expand the working operation.

$$\:\:{Expand}\:{the}\:{working}\:{operation}. \\ $$

Commented by som(math1967) last updated on 09/Mar/20

lnx^m y^n =ln(x+y)^(m+n) mlnx+nlny=(m+n)ln(x+y) (m/x) +(n/y)∙(dy/dx)=((m+n)/(x+y))(1+(dy/dx)) ((n/y) −((m+n)/(x+y)))(dy/dx)=(((m+n)/(x+y)) −(m/x)) ((nx−my)/(y(x+y))) ∙(dy/dx)=((nx−my)/(x(x+y))) ∴(dy/dx)=(y/x)

$${lnx}^{{m}} {y}^{{n}} ={ln}\left({x}+{y}\right)^{{m}+{n}} \\ $$$${mlnx}+{nlny}=\left({m}+{n}\right){ln}\left({x}+{y}\right) \\ $$$$\frac{{m}}{{x}}\:+\frac{{n}}{{y}}\centerdot\frac{{dy}}{{dx}}=\frac{{m}+{n}}{{x}+{y}}\left(\mathrm{1}+\frac{{dy}}{{dx}}\right) \\ $$$$\left(\frac{{n}}{{y}}\:−\frac{{m}+{n}}{{x}+{y}}\right)\frac{{dy}}{{dx}}=\left(\frac{{m}+{n}}{{x}+{y}}\:−\frac{{m}}{{x}}\right) \\ $$$$\frac{{nx}−{my}}{{y}\left({x}+{y}\right)}\:\centerdot\frac{{dy}}{{dx}}=\frac{{nx}−{my}}{{x}\left({x}+{y}\right)} \\ $$$$\therefore\frac{{dy}}{{dx}}=\frac{{y}}{{x}} \\ $$

Commented by niroj last updated on 09/Mar/20

thank you to all super heros.

$$\:\mathrm{thank}\:\mathrm{you}\:\mathrm{to}\:\mathrm{all}\:\mathrm{super}\:\mathrm{heros}. \\ $$

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