Calculating a Derivative

Properties 4.3.1  
Let $f$ and $g$ be differentiable functions. Then the following holds:

1. Sum Rule: $(f+g)'(x) = f'(x) + g'(x)$
2. Product Rule: $(f \cdot g)'(x) = f'(x)g(x) + f(x)g'(x)$
3. Chain Rule: $(g \circ f)'(x) = g'(f(x)) \cdot f'(x)$ if $g$ is defined in $f(x)$

Proof.

1. A straightforward calculation gives

   $$\begin{aligned}(f+g)'(x) &= \lim_{h \to 0}\frac{(f+g)(x+h)-(f+g... ..._{h \to 0}\frac{g(x+h)-g(x)}{h} \\ &= f'(x) + g'(x) \end{aligned}$$

   
   

2. By definition,

   $$\begin{aligned}(f \cdot g)'(x) &= \lim_{h \to 0}\frac{(f \cdot ... ...\ &= \lim_{h \to 0}\frac{(f(x+h)g(x+h)-f(x)g(x)}{h} \end{aligned}$$

   
   
   Inserting $f(x+h)g(x)-f(x+h)g(x)$ into the numerator gives

   $$(f \cdot g)'(x) = \lim_{h \to 0}\frac{(f(x+h)g(x+h)+f(x+h)g(x)-f(x+h)g(x)-f(x)g(x)}{h}$$ (4.3.3)

   
   
   and therefore

   $$(f \cdot g)'(x) = \lim_{h \to 0}f(x+h) \cdot \lim_{h \to 0}\frac{g(x+h)-g(x)}{h} + g(x) \cdot \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$ (4.3.4)

   
   
   As $f$ is differentiable, $f$ is continuous. Therefore $\lim_{h \to 0}f(x+h)=f(x)$, The other limits are $g'$ and $f'$, respectively, so (4.3.4) becomes

   $$(f \cdot g)'(x) = f(x) \cdot g'(x) \cdot f'(x) \cdot g(x)$$ (4.3.5)

   
   

3. By definition, and some playing around with elementary algebra, we have
   

   $$(g \circ f)' = \lim_{h \to 0}\frac{g(f(x+h))-g(f(x))}{h} \cr$$ (4.3.6)

   
   
   $f$ is differentiable, and therefore continuous (4.1.3), so we can insert

   $$\lim_{h \to 0}f(x+h) = \lim_{h \to 0}[f(x)+h]$$ (4.3.7)

   
   
   into (4.3.6) and we get
   

   $$(g \circ f)' = f'(x) \cdot \lim_{h \to 0}\frac{g(f(x)+h)-g(f(x))}{f(x)+h-f(x)} \cr$$ (4.3.8)

   
   

$\qedsymbol$

Remark 4.3.2 (Calculating a Derivative)  
Out of the derivatives for sum product and composition, quite a lot of other handy calculus rules can be derived. Two examples:

1. The well know formula

   $$\frac{d{(x^n)}}{dx} = n x^{n-1}$$ (4.3.9)

   
   
   can be derived using an induction argument. First, derive it for $n=2$ using the product rule:

   $$\frac{d{(x^2)}}{dx}= \frac{d{}}{dx} (x \cdot x) = x \cdot \frac{d{x}}{dx} + \frac{d{x}}{dx} \cdot x = 2x$$ (4.3.10)

   
   
   This can then be generalized to any $n$ with the following argument:

   $$\frac{d{(x^n)}}{dx}= \frac{d{}}{dx} (x \cdot x^{n-1}) = x \cdot \frac{d{(x^{n-1})}}{dx} + x^{n-1} = (n-1)xx^{n-2} + x^{n-1} = n x^{n-1}$$ (4.3.11)

   
   
   where we have used the induction hypothesis that the formula is valid for lower powers.
2. The so called ``Quotient Rule" can also then be obtained using the product rule and (4.3.9):

   $$\begin{aligned}\frac{d{}}{dx} \left( \frac{f}{g} \right) &= \fr... ...\\ &= \frac{\frac{d{f}}{dx}g-f\frac{d{g}}{dx}}{g^2} \end{aligned}$$

   
   

If using the sum rule, the product rule, or the chain rule are unpractical, then you can always resort to going back to the definition of derivative

$$\begin{aligned}\frac{d{}}{dx} \sin (x) &= \lim_{h \to 0}\frac{\... ...-1}{h} + \cos(x) \lim_{h \to 0}\frac{\sin(h)}{h} \\ \end{aligned}$$

Then note the following:

$$\lim_{h \to 0}\frac{\sin(h)}{h}=1 \hbox{\ \ and\ \ }\lim_{h \to 0}\frac{\cos(h)-1}{h}=0$$ (4.3.14)

So we can conclude

$$\frac{d{}}{dx}\sin(x) = \cos(x)$$ (4.3.15)

In practice, however, you will not do this, but you will use a table like the following

The First Derivative

Function	Derivative	Function	Derivative
$C$ (a constant)	0	$\vert x\vert$	$-1 (x<0),1 (x>0)$
$x$	$1$		undefined in 0
$x^n$	$nx^{n-1}$ (for any $n$)	$\arcsin(x)$	$\frac{1}{\sqrt{1-x^2}}$
$\frac{1}{x}$	$-\frac{1}{x^2}$	$\arccos(x)$	$-\frac{1}{\sqrt{1-x^2}}$
$\sqrt{x}$	$\frac{1}{2\sqrt{x}}$	$\arctan(x)$	$\frac{1}{1+x^2}$
$\root n \of x$	$\frac{1}{n \root n \of {x^{n-1}}}$	$\hbox{arccot}(x)$	$-\frac{1}{1+x^2}$
$e^x$	$e^x$	$\hbox{arcsec}(x)$	$\frac{1}{x\sqrt{x^2-1}}$
$a^x$	$a^x \ln(a)$	$\hbox{arccosec}(x)$	$-\frac{1}{x\sqrt{x^2-1}}$
$\ln(x)$	$\frac{1}{x}$	$\sinh(x)$	$\cosh(x)$
$\log_a (x)$	$\frac{1}{x} \log_a (x) = \frac{1}{x\ln(a)}$	$\cosh(x)$	$\sinh(x)$
$\sin(x)$	$\cos(x)$	$\tanh(x)$	$\frac{1}{\cosh^2(x)}$
$\cos(x)$	$-\sin(x)$	$\coth(x)$	$-\frac{1}{\sinh^2(x)}$
$\tan(x)$	$\frac{1}{\cos^2(x)} = \sec^2(x)$	$\hbox{arcsinh}(x)$	$\frac{1}{\sqrt{1+x^2}}$
$\cot(x)$	$-\frac{1}{\sin^2(x)} = \hbox{cosec}^2(x)$	$\hbox{arccosh}(x)$	$\frac{1}{\sqrt{x^2-1}}$
$\sec(x)$	$\tan(x)\sec(x)$	$\hbox{arctanh}(x)$	$\frac{1}{1-x^2}$
$\csc(x)$	$-\cot(x)\csc(x)$	$\hbox{arccoth}(x)$	$\frac{1}{1-x^2}$

Of the table above, you should make an honest attempt to memorize the left half and possible one of the inverse trigonometric functions. It is relatively easy to calculate the derivative of an inverse trigonometric function if you know the chain rule: First put $x = \sin(y)$ and note

$$\begin{aligned}\frac{d{x}}{dx} &= \cos(y) \frac{d{y}}{dx} \\ 1 ... ...x) \\ 1 &= \sqrt{1-x^2} \frac{d{}}{dx}\arcsin(x) \\ \end{aligned}$$

Now we take $\sqrt{1-x^2}$ to the other side, and we get

$$\frac{d{}}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$$