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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. |
$f'(x)$ = $\lim_{h \rightarrow 0}$$\frac{f(x + h) - f(x)}{h}$
Solved Example
$f(x)$ = $x^2$ + 5x
Solution:
$ƒ(x)$ = $x^2$ + 5x
ƒ(x+h) = $(x+h)^2$ + 5(x+h)
= $x^2 + 2hx + h^2 + 5x + 5h$
Apply the definition of derivative
$ƒ'(x)$ = $\lim_{h \rightarrow 0}$ $\frac{f(x + h) - f(x)}{h}$
So, $ƒ'(x)$ = $\lim_{h \rightarrow 0}$$\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.
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$ = -$\frac{1}{1 + x^2}$ |
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}$$\frac{(3x^2 + 1)}{(x^2 - 1)}$ = $\frac{(x^2 - 1)(6x) - (3x^2 + 1)(2x)}{(x^2 - 1)^2}$
$\frac{d}{dx}$ $\frac{3x^2 + 1}{x^2 - 1}$ = $\frac{6x^3 - 6x - 6x^3 - 2x}{(x^2 - 1)^2}$
$\frac{d}{dx}$ $\frac{3x^2 + 1}{x^2 - 1}$ = $\frac{-8x}{(x^2 - 1)^2}$
Solved Examples
Solution:
$ƒ(x)$ = $\frac{1}{x + 2}$
so $ƒ(x + h)$ = $\frac{1}{x + h + 2}$
by definition of derivative
$ƒ’(x)$ = $\lim_{h \rightarrow 0}$$\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
$ƒ’(x)$ = $\frac{1}{(x + 2)(x + 2)}$
= $\frac{1}{(x + 2)^2}$
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
$ƒ'(x)$ = $\frac{1}{2 \sqrt{x + 4}}$
$ƒ(x)$ = ($x^2 + x$) (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)
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 – $sec^2 x$
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
$V(t) = t^3 - 6t^2 + 35$
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
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 + $\frac{1}{\sqrt{(1 - x^2)}}$
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)
= $\frac{5}{\sqrt{1 - 25x^2 - 20x - 4}}$
$\frac{5}{\sqrt{-25x^2 - 20x - 3}}$
