A derivative is the instantaneous rate of change of a function. For example while finding rates of change, the instantaneous rate of change of a function f(x) when x = a is equal to the derivative of f at a.

If f(x) = y and for a small change in x say $\partial$x , the function changes by $\partial$y. If the
$\lim_{\partial x \rightarrow 0}$ $\frac{\partial y}{\partial x}$ exists then it is called as the derivative of the function.
Derivatives have their own notation. The derivative of a function f at a number a is denoted by f'(a) and is given by:

$f'(a) = $$\lim_{h \rightarrow 0}$$\frac{f(a+h) - f(a)}{h}$ = $\lim_{x \rightarrow a}$$\frac{f(x) - f(a)}{x - a}$

The most common notation to denote derivative is $f'(x)$. Some other notations are
$f'(x) = y'$ = $\frac{dy}{dx}$ = $\frac{d}{dx}$ $f(x)$

The derivative of a function f(x) with respect to x is defined as

$f'(x)$ = $\lim_{h \rightarrow 0}$$\frac{f(x + h) - f(x)}{h}$

Solved Example

Question: Solve the derivative of the given function by using derivative definition.
$f(x)$ = $x^2$ + 5x
Solution:
 
$ƒ(x)$ = $x^2$ + 5x
ƒ(x+h)  = $(x+h)^2$ + 5(x+h)
              = $x^2 + 2hx + h^2 + 5x + 5h$
              
Apply the definition of derivative

    $ƒ'(x)$ = $\lim_{h \rightarrow 0}$ $\frac{f(x + h) - f(x)}{h}$
    
         So, $ƒ'(x)$ = $\lim_{h \rightarrow 0}$$\frac{x^2 + 2hx + h^2 + 5x + 5h - (x^2 + 5x)}{h}$
        
$\lim_{h \rightarrow 0}$$\frac{2hx + h^2 + 5h}{h}$

Take h common from numerator.

= $\lim_{h \rightarrow 0}$$\frac{h(2x + h + 5)}{h}$

Cancel out h and simplify

$ƒ'(x)$ = $\lim_{h \rightarrow 0}$ 2x + h + 5

         Plug in the limit h=0,
        
$ƒ‘(x)$ = 2x + 0 + 5 = 2x + 5.
 

There are rules for finding derivative of the functions

1. Derivative of a constant = 0
if ƒ (x) = c then ƒ’(x) = 0

2. Power rule
$ƒ(x)$ = $x^n$
then $ƒ'(x)$ = n $x^{n-1}$

3. Scalar Multiple $\frac{d}{dx}$ c $ƒ(x)$ = c $ƒ'(x)$ where c is a constant

4. Sum $\frac{d}{dx}$ (u + v) = u' + v'

5. Difference $\frac{d}{dx}$ (u - v) = $u' - v'$

6. Product rule: $\frac{d}{dx}$ ($u \times v$) = $u \times v' + v \times u'$

7. Quotient rule: $\frac{d}{dx} \frac{u}{v}$ = $\frac{v \times u' - u \times v'}{v^2}$

8. Reciprocal rule $\frac{d}{dx}$ ($\frac{1}{v}$) = $\frac{v'}{v^2}$

9. chain Rule $\frac{d}{dx}$ ƒ(g(x)) = ƒ'(g(x)) $\times$ g'(x)
u and v are two different differentiable functions. → Read More
Power of x
$\frac{d}{dx}$c = 0 $\frac{d}{dx}$x = 1 $\frac{d}{dx}$$x^n = n x^{n-1}$

Exponential or logarithmic functions
$\frac{d}{dx}$$e^x$ = $e^x$ $\frac{d}{dx}$$b^x$ = $b^x In(b)$ $\frac{d}{dx}$$In(x)$ = $\frac{1}{x}$

Trigonometric functions
$\frac{d}{dx}$$sin x$ = $cos x$ $\frac{d}{dx}$$csc x$ = $- csc x cot x$
$\frac{d}{dx}$$cos x$ = $-sin x$ $\frac{d}{dx}$$sec x$ = $sec x tan x$
$\frac{d}{dx}$$tan x$ = $sec^2 x$ $\frac{d}{dx}$$cot x$ = $-csc^2 x$

Inverse Trigonometric functions
$\frac{d}{dx}$$arc sin x$ = $\frac{1}{\sqrt{1 - x^2}}$ $\frac{d}{dx}$$arccsc x$ = $\frac{-1}{|x| \sqrt{x^2 - 1}}$
$\frac{d}{dx}$$arccos x$ = $\frac{-1}{\sqrt{1 - x^2}}$ $\frac{d}{dx}$$arcsec x$ = $\frac{1}{|x| \sqrt{x^2 - 1}}$
$\frac{d}{dx}$$tan x$ = $\frac{1}{1 + x^2}$ $\frac{d}{dx}$$arccot x$ = -$\frac{1}{1 + x^2}$
Derivative of fraction can be computed by applying quotient rule.
Suppose the function is $\frac{3x^2 + 1}{x^2 - 1}$ then we can apply quotient rule since it is in the form u/v where both u and v are functions of x

u = 3$x^2$ + 1

v = $x^2$ - 1

$\frac{d}{dx}$ ($\frac{u}{v}$) = $\frac{v \times u' - u \times v'}{v^2}$

u' = $\frac{d}{dx}$(3$x^2$ + 1) = $\frac{d}{dx}$ $3x^2$ + $\frac{d}{dx}$ 1

= 3 $\times$ 2x + 0

u' = 6x

v = $x^2 - 1$

v' = $\frac{d}{dx}$ ($x^2 - 1$)

= $\frac{d}{dx}$ $x^2$ -$\frac{d}{dx}$ 1

= 2x - 0

v' = 2x

$\frac{d}{dx}$$\frac{(3x^2 + 1)}{(x^2 - 1)}$ = $\frac{(x^2 - 1)(6x) - (3x^2 + 1)(2x)}{(x^2 - 1)^2}$

$\frac{d}{dx}$ $\frac{3x^2 + 1}{x^2 - 1}$ = $\frac{6x^3 - 6x - 6x^3 - 2x}{(x^2 - 1)^2}$

$\frac{d}{dx}$ $\frac{3x^2 + 1}{x^2 - 1}$ = $\frac{-8x}{(x^2 - 1)^2}$

Solved Examples

Question 1: $ƒ(x)$ = $\frac{1}{x + 2}$
Solution:
 
$ƒ(x)$ = $\frac{1}{x + 2}$

so $ƒ(x + h)$ = $\frac{1}{x + h + 2}$

by definition of derivative

$ƒ’(x)$ = $\lim_{h \rightarrow 0}$$\frac{f(x + h) - f(x)}{h}$

=$\lim_{h \rightarrow 0}$$\frac{\frac{1}{x + h + 2}- \frac{1}{x+2}}{h}$

Let us solve the numerator first

$\frac{1}{x + h + 2}$ - $\frac{1}{x + 2}$

LCD = $(x+h+2)(x+2)$

so we have

$\frac{x + 2}{(x + h + 2)(x + 2)}$ - $\frac{x + h + 2}{(x + 2)(x + h + 2)}$

$\frac{h}{(x + h + 2)(x + 2)}$

$ƒ’(x)$ = $\lim_{h \rightarrow 0}$$\frac{h/(x + h + 2)(x + 2)}{h}$

= $\lim_{h \rightarrow 0}$ $\frac{1}{(x + h + 2)(x+2)}$

plug in the limit h = 0

$ƒ’(x)$ = $\frac{1}{(x + 2)(x + 2)}$

= $\frac{1}{(x + 2)^2}$
 

Question 2: $ƒ(x)$ = $\sqrt{(x + 4)}$
Solution:
 
$ƒ(x)$ = $\sqrt{(x + 4)}$

so $ƒ(x + h)$ = $\sqrt{(x + h + 4)}$

Apply the definition of derivative

$ƒ‘(x)$ = $\lim_{h \rightarrow 0}$$\frac{(\sqrt{(x + h + 4)}) - (\sqrt{(x + 4)})}{h}$

Multiply with conjugate of ($\sqrt{(x + h + 4)}$) - $(\sqrt{(x + 4)}$

= $\lim_{h \rightarrow 0}$$\frac{(\sqrt{x + h + 4}) - (\sqrt{x + 4})}{h}$ $\times$ $\frac{(\sqrt{x + h + 4}) + (\sqrt{x + 4})}{(\sqrt{x + h + 4}) + (\sqrt{x + 4})}$

= $\lim_{h \rightarrow 0}$$\frac{(\sqrt{x + h + 4})^2 - (\sqrt{x + 4})^2}{h((\sqrt{x + h + 4}) + \sqrt{x + 4}))}$

= $\lim_{h \rightarrow 0}$$\frac{x + 4 + h - x - 4}{h((\sqrt{x + h + 4}) + \sqrt{x + 4}))}$


= $\lim_{h \rightarrow 0}$$\frac{h}{h((\sqrt{x + h + 4}) + \sqrt{x + 4}))}$

Cancel out h

$ƒ’(x)$ == $\lim_{h \rightarrow 0}$$\frac{1}{h((\sqrt{x + h + 4}) + \sqrt{x + 4}))}$

Plugging in h = 0 we get

$ƒ'(x)$ = $\frac{1}{2 \sqrt{x + 4}}$
 

Question 3: Find derivative by applying rules

$ƒ(x)$ = ($x^2 + x$) (sin x)
Solution:
 
Here given function is product of two different functions

u= $x^2$ + x and v = sin x

Apply the product rule

$\frac{d}{dx}$ (u $\times$ v) = u $\times$ v' + v $\times$ u'

$\frac{d}{dx}$ ($x^2 + x$)(sin x) = ($x^2 + x$) $\frac{d}{dx}$ sinx + sinx $\times$ $\frac{d}{dx}$ ($x^2 + x$)

= ($x^2 + x$) cosx + sinx $\times$(2x + 1)
 

Question 4: ƒ(x) = ln(2x) + 5x – tan x
Solution:
 
ƒ’(x) = $\frac{d}{dx}$ ln(2x) + $\frac{d}{dx}$ 5x - $\frac{d}{dx}$ tanx

= $\frac{1}{2x}$ $\frac{d}{dx}$ 2x + 5 - $sec^2 x$

ƒ’(x) = $\frac{1}{x}$ + 5 – $sec^2 x$
 

Question 5: Find slope of the tangent to the curve ƒ(x) = 15 - $2x^2$ at x=1, hence find the equation of the tangent line.
Solution:
 
Slope of the tangent at point x = a is given by

ƒ’(x) = $\lim_{x \rightarrow a}$ $\frac{f(x) - f(a)}{x - a}$

= $\lim_{x \rightarrow 1}$ $\frac{15 - 2x^2 - (15 - 2(1)^2)}{x - 1}$

= $\lim_{x \rightarrow 1}$ $\frac{2 - 2x^2}{x - 1}$

= $\lim_{x \rightarrow 1}$ $\frac{2(1 - x^2)}{x - 1}$

= $\lim_{x \rightarrow 1}$ -2(1 + x)

m = $f’(x)$ = -4

Equation of the tangent line at x = 1, the point (x, ƒ(x)) will be

m = -4, (1, 13)

$y - y_0$ = m($x - x_0$)

y - 13 = -4(x-1)

y - 13 = -4x + 4

y = -4x + 17
 

Question 6: Suppose that the amount of air in a balloon after t hours is given by

$V(t) = t^3 - 6t^2 + 35$

Estimate the instantaneous rate of change of the volume after 5 hours.
Solution:
 
To find the instantaneous change we can take the derivative of the function at t=5.

So instantaneous rate of change of volume = $\frac{dv}{dt}$

$\frac{dv}{dt}$ = $\frac{d}{dt}$ ($t^3 - 6t^2 + 35$)

= $3t^2 - 6 \times 2t + 0$

= $3t^2 - 12t$

At t=5 $\frac{dv}{dt}$ is

$3(5)^2 - 12 \times 5$

= 15
 

Question 7: Differentiate $x^2 + 2^x + arcsin x$
Solution:
 
Apply the rules from the derivative table.

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

$\frac{d}{dx}$ $b^x$ = $b^x$ ln(b)

ƒ(x) = $x^2 + 2^x + arcsin x$

ƒ'(x) = $\frac{d}{dx}$ $x^2 + 2^x + arcsin x$

ƒ'(x) = 2x + $2^x$ ln 2 + $\frac{1}{\sqrt{(1 - x^2)}}$
 

Question 8: Differentiate arcsin(5x + 2)
Solution:
 
f(x) = arcsin(5x + 2)

f’(x) = $\frac{d}{dx}$ arcsin(5x + 2)

= $\frac{1}{\sqrt{(1 - (5x + 2)^2}}$ $\times$ $\frac{d}{dx}$(5x + 2)

= $\frac{5}{\sqrt{1 - 25x^2 - 20x - 4}}$

$\frac{5}{\sqrt{-25x^2 - 20x - 3}}$
 

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