Mohon bantuannya Ngasal jawab/Bahasa alien, auto report sampe ke akar akar

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

Mohon bantuannya

Ngasal jawab/Bahasa alien, auto report sampe ke akar akar
Mohon bantuannya
Ngasal jawab/Bahasa alien, auto report sampe ke akar akar

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\large\text{$\begin{aligned}f_x(4,3)+f_y(4,3)=\boxed{\bf15}\end{aligned}$}$
 

Pembahasan

Turunan Parsial

Diberikan: $f(x,y)=3xe^{y-3}$

Maka,

$\begin{aligned}&f_x(4,3)+f_y(4,3)\\&{=\ }f_x(x,y)+f_y(x,y)&_{(x=4,y=3)}\\&{=\ }\frac{\partial}{\partial x}\left(3xe^{y-3}\right)+\frac{\partial}{\partial y}\left(3xe^{y-3}\right)&_{(x=4,y=3)}\\&{=\ }3e^{y-3}\left[\frac{\partial}{\partial x}(x)\right]+3x\left[\frac{\partial}{\partial y}\left(e^{y-3}\right)\right]&_{(x=4,y=3)}\\&{=\ }3e^{y-3}(1)+3x\left[e^{y-3}\cdot\frac{\partial}{\partial y}(y-3)\right]&_{(x=4,y=3)}\\&{=\ }3e^{y-3}+3x\left[e^{y-3}\cdot1\right]&_{(x=4,y=3)}\end{aligned}$
$\begin{aligned}&{=\ }3(1+x)e^{y-3}&_{(x=4,y=3)}\\&{=\ }3(1+4)e^{3-3}=15e^{0}\\&{=\ }\boxed{\bf15}\end{aligned}$
$\blacksquare$

Dengan cara limit

$\begin{aligned}f_x(4,3)&=\frac{\partial f(4,3)}{\partial x}\\&=\lim_{h\to0}\frac{f(4+h,3)-f(4,3)}{h}\\&=\lim_{h\to0}\frac{3(4+h)e^{3-3}-3(4)e^{3-3}}{h}\\&=\lim_{h\to0}\frac{(12+3h)e^{0}-12e^{0}}{h}\\&=\lim_{h\to0}\frac{12+3h-12}{h}\\&=\lim_{h\to0}\frac{3h}{h}=\lim_{h\to0}3\\f_x(4,3)&=\bf3\end{aligned}$

$\begin{aligned}f_y(4,3)&=\frac{\partial f(4,3)}{\partial y}\\&=\lim_{h\to0}\frac{f(4,3+h)-f(4,3)}{h}\\&=\lim_{h\to0}\frac{3(4)e^{3+h-3}-12}{h}\\&=\lim_{h\to0}\frac{12e^{h}-12}{h}\\&=\lim_{h\to0}\frac{12\left(e^{h}-1\right)}{h}\\&=12\cdot\lim_{h\to0}\frac{e^{h}-1}{h}\\&\ \to\textsf{\sf Dalil\ L'H\^{o}pital}\\&=12\cdot\lim_{h\to0}\frac{\frac{d}{dh}\left(e^{h}-1\right)}{\frac{d}{dh}(h)}\\&=12\cdot\lim_{h\to0}\frac{e^{h}-0}{1}\\&=12\cdot\lim_{h\to0}e^{h}\end{aligned}$
$\begin{aligned}&=12\cdot e^0=12\cdot1\\f_y(4,3)&=\bf12\end{aligned}$