Calculus 12: Basic Differentiation Formulas

 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 $x$. Note that $\Delta x$ could be positive or negative. Denote by $\Delta y$ the difference $f(x+\Delta x)-f(x)$. $\Delta y$ is called an increment of $y$. Hence, the average rate of change of $y$ with respect to $x$ in the interval $[x,x+\Delta x]$ is the difference quotient $$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}.$$  In Gottfried Leibniz's viewpoint, one could think of $\Delta x$ as becoming infinitesimal. The resulting quantity is denoted by $dx$. When $\Delta x$ becomes the infinitesimal $dx$, $\Delta y$ simultaneously becomes the infinitesimal $dy$. The infinitesimals $dx$ and $dy$ are called differentials. Hence the ratio $\frac{\Delta y}{\Delta x}$ becomes $\frac{dy}{dx}$ accordingly, and it is exactly equal to $f'(x)$. The quantity $\frac{dy}{dx}$ can be viewed as the ratio of differentials or as s synonym for $f'(x)$.

Leibniz Notation: If $y=f(x)$, the derivative $f'(x)$ can be written $$\frac{dy}{dx},\frac{df(x)}{dx},\ \mbox{or}\ \frac{d}{dx}f(x).$$ This is just a notation and does not represent a division. Using Leibniz notaion, the value $f'(a)$ of $f'(x)$ at a specific point $x=a$ can be written $$\left.\frac{dy}{dx}\right|_{x=a}\ \mbox{or}\ \left.\frac{df(x)}{dx}\right|_{x=a}.$$

Calculating derivatives using the definition can be really laborious. In actual practice, special rules and formulas are derived for differentiating certain types of functions. The following are such rules and they can be proved straightforwardly by the definition of the derivative.

Theorem. Let $c$ be a constant, and $f(x)$ and $g(x)$ be two differentiable functions of $x$.

1. $\displaystyle\frac{dc}{dx}=0$
2. $\displaystyle\frac{d(cf(x))}{dx}=c\frac{df(x)}{dx}$
3. $\displaystyle\frac{d[f(x)+g(x)]}{dx}=\frac{df(x)}{dx}+\frac{dg(x)}{dx}$
   

The converse of the first rule is also true, namely if $f'(x)=0$ for all $x$ in the domain then $f(x)$ is a constant function. This can be proved using the Mean Value Theorem which will be studied later.

Lemma. [Binomial Theorem] \begin{align*}(a+b)^n&=\begin{pmatrix}n\\0\end{pmatrix}a^nb^0+ \begin{pmatrix}n\\1\end{pmatrix}a^{n-1}b+\begin{pmatrix}n\\2\end{pmatrix}a^{n-2}b^2+\cdots+\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k+\\&\cdots+\begin{pmatrix}n\\n-1\end{pmatrix}ab^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}a^0b^n\\&=a^n+\begin{pmatrix}n\\1\end{pmatrix}a^{n-1}b+\begin{pmatrix}n\\2\end{pmatrix}a^{n-2}b^2+\cdots+\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k+\\&\cdots+\begin{pmatrix}n\\n-1\end{pmatrix}ab^{n-1}+b^n,\end{align*} where $$\begin{pmatrix}n\\k\end{pmatrix}=\frac{n!}{k!(n-k)!}.$$ $\begin{pmatrix}n\\k\end{pmatrix}$ is also denoted by $n{\mathrm C}k$. The binomial coefficients $\begin{pmatrix}n\\k\end{pmatrix}$ can be also easily obtained by Pascal's triangle. For details see here and here.

Theorem. [Power Rule] $\displaystyle\frac{dx^n}{dx}=nx^{n-1}$

Proof. Let $y=x^n$. Then by the Binomial Theorem \begin{align*}\frac{dx^n}{dx}&=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}\\&=\lim_{\Delta x\to 0}\frac{(x+\Delta x)^n-x^n}{\Delta x}\\&=\lim_{\Delta x\to 0}\frac{\left[x^n+nx^{n-1}\Delta x+\frac{n(n-1)}{2!}x^{n-2}(\Delta x)^2+\cdots+(\Delta x)^n\right]-x^n}{\Delta x}\\&=\lim_{\Delta x\to 0}\left[nx^{n-1}+\frac{n(n-1)}{2!}x^{n-2}\Delta x+\cdots+(\Delta x)^{n-1}\right]\\&=nx^{n-1}.\end{align*}

Example. If $y=3x^5$, then by Property 2 of the first theorem and Power Rule (the preceding theorem), we have \begin{align*}\frac{dy}{dx}&=3\frac{dx^5}{dx}\\&=3(5)x^{5-1}\\&=15x^4.\end{align*}

Remark. In the preceding theorem, the power rule is established only for the case when $n$ is a positive integer. The formula is indeed valid for all real $n$'s.

Example. If $y=8x^{-\frac{3}{4}}$, then we have $$\frac{dy}{dx}=8\left(-\frac{3}{4}\right)x^{-\frac{3}{4}-1}=-6x^{-\frac{7}{4}}.$$

Using the first theorem and the preceding theorem (power rule), we can now find the derivative of any polynomial function.

Example. If $y=2x^4-x^3-2x+7$, then \begin{align*}\frac{dy}{dx}&=\frac{d(2x^4)}{dx}-\frac{d(x^3)}{dx}-2\frac{dx}{dx}+\frac{d(7)}{dx}\\&=8x^3-3x^2-2.\end{align*}

Example. The function $f(x)=\displaystyle\frac{3x^3-4}{x^2}$ can be written in the form to which we can apply the power rule to find its derivative. $$f(x)=\frac{3x^3-4}{x^2}=\frac{3x^3}{x^2}-\frac{4}{x^2}=3x-4x^{-2}.$$ Hence we have $$f'(x)=\frac{d(3x)}{dx}-\frac{d(4x^{-2})}{dx}=3+8x^{-3}.$$

Example. The function $y=\root 3\of{x^2}-3\root 3\of{x}-5$ can be also written in the form to which we can apply the power rule to find its derivative. \begin{align*}y&=\root 3\of{x^2}-3\root 3\of{x}-5\\&=(x^2)^{\frac{1}{3}}-3x^{\frac{1}{3}}-5\\&=x^{\frac{2}{3}}-3x^{\frac{1}{3}}-5.\end{align*} Hence, $$\frac{dy}{dx}=\frac{2}{3}x^{-\frac{1}{3}}-x^{-\frac{2}{3}}.$$