Genel olarak şu kuralı arıyoruz;
$U(x_1(t),x_2(t),....,x_i(t))$ diye bir türevlenebilen fonksiyon tanımlarsak;
$$\dfrac{dU}{dt}=\dfrac{\partial U}{\partial x_1}\dfrac{dx_1}{dt}+\dfrac{\partial U}{\partial x_2}\dfrac{dx_2}{dt}+...+\dfrac{\partial U}{\partial x_i}\dfrac{dx_i}{dt}$$
olur ama neden?
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$x,y$ fonksiyonları $t$'ye bağlı olsun
$f(x(t),y(t))$ diye türevlenebilen bir fonksiyon tanımlıyoruz:
$$\dfrac{df}{dt}=\dfrac{\partial f}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial f}{\partial y}\dfrac{dy}{dt}$$
Olduğunu ispatlayalım
$G(t)=f(x(t),y(t))$ olsun;
$$\begin{align}\frac{\triangle G}{\triangle t}&=\frac{f(x(t+h),y(t+h))-f(x(t),y(t))}h\\&=\frac{f(x(t+h),y(t+h))\color{#3388dd}{-f(x(t+h),y(t))+f(x(t+h),y(t))}-f(x(t),y(t))}h\\&=\color{red}{\underbrace{\color{black}{\frac{f(x(t+h),y(t+h))-f(x(t+h),y(t))}h}}_{A}}+\color{red}{\underbrace{\color{black}{\frac{f(x(t+h),y(t))-f(x(t),y(t))}h}}_{B}}\end{align}$$
$\lim\limits_{h\to 0}\dfrac{\triangle G}{\triangle t}=\dfrac{dG}{dt}$
olduğundan
$$\begin{align}A&=\frac{f(x(t+h),y(t+h))-f(x(t+h),y(t))}h\\&=\frac{f(x(t+h),y(t+h))-f(x(t+h),y(t))}{\color{#3388dd}{y(t+h)-y(t)}}\frac{\color{#3388dd}{y(t+h)-y(t)}}h\\&\to\lim\limits_{h\to 0}A=\frac{\partial f}{\partial y}\frac{dy}{dt}\end{align}$$
$$\begin{align}B&=\frac{f(x(t+h),y(t))-f(x(t),y(t))}h\\&=\frac{f(x(t+h),y(t))-f(x(t),y(t))}{\color{#3388dd}{x(t+h)-y(t)}}\frac{\color{#3388dd}{x(t+h)-y(t)}}h\\&\to\lim\limits_{h\to 0}B=\frac{\partial f}{\partial x}\frac{dx}{dt}\end{align}$$
$\dfrac{dG}{dt}=\lim\limits_{h\to 0}(A+B)=\dfrac{\partial f}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial f}{\partial x}\dfrac{dx}{dt}$
İspat biter. $\Box$