Monthly Archives: March 2011

Product Rule: Let $u=f(x)$ and $v=g(x)$ be differentiable functions. Then $$(fg)’(x)=f(x)g’(x)+f’(x)g(x)$$ or $$\frac{d(uv)}{dx}=u\frac{dv}{dx}+\frac{du}{dx}v.$$ Proof. \begin{eqnarray*}(fg)’(x)&=&\lim_{\Delta x\to 0}\frac{fg(x+\Delta x)-fg(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}g(x+\Delta x)+f(x)\lim_{\Delta x\to 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}\\&=&f’(x)g(x)+f(x)g’(x).\end{eqnarray*} … Continue reading →

There is a close relatioship between continuity and differentiability, namely Theorem 18. If $f’(x_0)$ exists then $f(x)$ is continuous at $x_0$; i.e. $\displaystyle\lim_{x\to x_0}f(x)=f(x_0)$. However the converse need not be true. Proof. \begin{eqnarray*}\lim_{x\to x_0}[f(x)-f(x_0)]&=&\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)\\&=&f’(x)\cdot 0\\&=&0.\end{eqnarray*} Example. [A Counterexample for … Continue reading →

Let us recall the definition of the derivative $$f’(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$ Replace $h$ by $\Delta x$. Then $f’(x)$ is rewritten as $$f’(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}.$$ In mathematics, $\Delta$ often means an increment. So $\Delta x$ means an increment of $x$. … Continue reading →

Let us assume that a particle is moving along a straight line and that the function $s=f(t)$ describes the position of moving particle at the time $t$. In physics, such a function $s=f(t)$ is called a motion. Suppose the particle … Continue reading →