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Diketahui f(x, y) = In √x+y/x-y Tunjukkan bahwa fxy =

Berikut ini adalah pertanyaan dari sabiraarrasyida16 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Diketahui f(x, y) = In √x+y/x-y Tunjukkan bahwa fxy = fyx.​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

\boxed{\,\begin{aligned}\vphantom{\Bigg|}&f_{xy}=f_{yx}=-\frac{x^2+y^2}{{\left(x^2-y^2\right)}^2}\\\vphantom{\bigg|}&\therefore\ f_{xy}=f_{yx}{\sf\ terbukti!}\end{aligned}\,}
 

Penjelasan dengan langkah-langkah:

Diketahui:
\begin{aligned}f(x,y)&=\ln\left(\sqrt{\frac{x+y}{x-y}}\,\right)\end{aligned}

Kita akan menunjukkan/membuktikan bahwa:
f_{xy}=f_{yx}

PEMBUKTIAN

\begin{aligned}f(x,y)&=\ln\left(\sqrt{\frac{x+y}{x-y}}\right)\\&=\frac{1}{2}\ln\left(\frac{x+y}{x-y}\right)\\\end{aligned}

Turunan parsial pertama

\begin{aligned}f_{x}&=\frac{\partial f}{\partial x}\\&=\frac{\partial}{\partial x}\left(\frac{1}{2}\ln\left(\frac{x+y}{x-y}\right)\right)\\&=\frac{1}{2}\cdot\frac{\partial}{\partial x}\ln\left(\frac{x+y}{x-y}\right)\\&=\frac{1}{2}\cdot\frac{1}{\frac{x+y}{x-y}}\cdot\frac{\partial }{\partial x}\left(\frac{x+y}{x-y}\right)\end{aligned}
\begin{aligned}&=\frac{\cancel{x-y}}{2(x+y)}\cdot\frac{\left[\frac{\partial }{\partial x}(x+y)\right](x-y)-(x+y)\left[\frac{\partial }{\partial x}(x-y)\right]}{(x-y)^{\cancel{2}}}\\&=\frac{1}{2(x+y)}\cdot\frac{(x-y)-(x+y)}{(x-y)}\\&=\frac{1}{\cancel{2}(x+y)}\cdot\frac{-\cancel{2}y}{(x-y)}\\&=-\frac{y}{(x+y)(x-y)}\\f_{x}&=-\frac{y}{x^2-y^2}\end{aligned}

\begin{aligned}f_{y}&=\frac{\partial f}{\partial y}\\&=\frac{\partial}{\partial y}\left(\frac{1}{2}\ln\left(\frac{x+y}{x-y}\right)\right)\\&=\frac{1}{2}\cdot\frac{\partial}{\partial y}\ln\left(\frac{x+y}{x-y}\right)\\&=\frac{1}{2}\cdot\frac{1}{\frac{x+y}{x-y}}\cdot\frac{\partial }{\partial y}\left(\frac{x+y}{x-y}\right)\end{aligned}
\begin{aligned}&=\frac{\cancel{x-y}}{2(x+y)}\cdot\frac{\left[\frac{\partial }{\partial y}(x+y)\right](x-y)-(x+y)\left[\frac{\partial }{\partial y}(x-y)\right]}{(x-y)^{\cancel{2}}}\\&=\frac{1}{2(x+y)}\cdot\frac{(x-y)+(x+y)}{(x-y)}\\&=\frac{1}{\cancel{2}(x+y)}\cdot\frac{\cancel{2}x}{(x-y)}\\&=\frac{x}{(x+y)(x-y)}\\f_{y}&=\frac{x}{x^2-y^2}\end{aligned}

Turunan Parsial Kedua

\begin{aligned}f_{xy}&=\frac{\partial f_x}{\partial y}\\&=\frac{\partial}{\partial y}\left(-\frac{y}{x^2-y^2}\right)\\&=-\frac{\partial}{\partial y}\left(\frac{y}{x^2-y^2}\right)\\&=-\frac{\left(\frac{\partial}{\partial y}y\right)\left(x^2-y^2\right)-y\left[\frac{\partial}{\partial y}\left(x^2-y^2\right)\right]}{{\left(x^2-y^2\right)}^2}\\&=-\frac{x^2-y^2-y(-2y)}{{\left(x^2-y^2\right)}^2}\end{aligned}
\begin{aligned}&=-\frac{x^2-y^2+2y^2}{{\left(x^2-y^2\right)}^2}\\f_{xy}&=-\frac{x^2+y^2}{{\left(x^2-y^2\right)}^2}\\\end{aligned}
\begin{aligned}f_{yx}&=\frac{\partial f_y}{\partial x}\\&=\frac{\partial}{\partial x}\left(\frac{x}{x^2-y^2}\right)\\&=\frac{\left(\frac{\partial}{\partial x}x\right)\left(x^2-y^2\right)-x\left[\frac{\partial}{\partial x}\left(x^2-y^2\right)\right]}{{\left(x^2-y^2\right)}^2}\\&=\frac{x^2-y^2-x(2x)}{{\left(x^2-y^2\right)}^2}\end{aligned}
\begin{aligned}&=\frac{x^2-2x^2-y^2}{{\left(x^2-y^2\right)}^2}\\&=\frac{-x^2-y^2}{{\left(x^2-y^2\right)}^2}=\frac{-\left(x^2+y^2\right)}{{\left(x^2-y^2\right)}^2}\\f_{yx}&=-\frac{x^2+y^2}{{\left(x^2-y^2\right)}^2}\\\end{aligned}

Kesimpulan

\boxed{\,\begin{aligned}\vphantom{\Bigg|}&f_{xy}=f_{yx}=-\frac{x^2+y^2}{{\left(x^2-y^2\right)}^2}\\\vphantom{\bigg|}&\therefore\ f_{xy}=f_{yx}{\sf\ terbukti!}\end{aligned}\,}
\blacksquare

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Last Update: Fri, 13 Jan 23
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