Theorem 3.3.1.
\(sin'(x)=cos(x), cos'(x)=-sin(x), (e^x)'=e^x\) and \(ln'(x)=1/x\)
Section 3.3 Basic Properties
So, in previous sections, we figured out a few derivatives like \(k'=0\text{,}\) \(x'=1\) and \((x^2)'=2x\text{.}\) Now what? Well, careful application of the definition of derivative would also give the following
Even better, derivative is as robust a concept as limit in that it yields the following basic properties theorem.
Theorem 3.3.2.
- If \(r\) and \(s\) are any numbers and \(f\) and \(g\) are differentiable at \(x\text{,}\) then \((rf+sg)'(x)=rf'(x)+sg'(x)\text{.}\)
- If \(f\) and \(g\) are differentiable at \(x\text{,}\) then \((fg)'(x)=f'(x)g(x)+f(x)g'(x)\text{.}\)
- If \(f\) and \(g\) are differentiable at \(x\) and \(g(x)\ne 0\text{,}\) then \((\frac{f}{g})'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}\text{.}\)
Perhaps the first obvious thing we can prove from these basic facts is that \((x^3)'=3x^2\text{.}\) How? Do you see it? Well, notice that \(x^3=x^2\) times \(x\text{.}\) So, the product property 2 of derivatives in the theorem above applies here giving \((x^3)'=(x^2x)'=(x^2)'x+x^2x' = 2xx + x^2= 3x^2\text{.}\) Similarly we could get \((x^4)'=4x^3\text{.}\) Do you see a pattern: \((x^2)'=2x\text{,}\) \((x^3)'=3x^2\text{,}\) \((x^4)'=4x^3\text{,}\) etc. In greatest generality, we have
Theorem 3.3.3.
If \(x^{r-1}\) is defined on a given open interval, then \((x^r)'=rx^{r-1}\) there
More dramatic is that using the few derivatives we know and this basic properties theorem, we are now able to differentiate a large swath of functions. Let’s throw one together like
\begin{equation*}
f(x)=\frac{sin(x)}{x^2+1}
\end{equation*}
We now can compute
\begin{equation*}
\begin{array}{llll}
f'(x) \amp = \amp (\frac{sin(x)}{x^2+1})' \amp (\text{from definition of f}) \\
\amp = \amp \frac{sin'(x)(x^2+1)-sin(x)(x^2+1)'}{(x^2+1)^2} \amp (\text{the quotient property 3 of derivative}) \\
\amp = \amp \frac{cos(x)(x^2+1)-sin(x)((x^2)'+1')}{(x^2+1)^2} \amp (\text{known derivatives + sum property 1 of derivative}) \\
\amp = \amp \frac{cos(x)(x^2+1)-sin(x)(2x)}{(x^2+1)^2} \amp (\text{known derivative})
\end{array}
\end{equation*}
