Differentiation

A function f(x) defined in the open interval (a, b) is said to be differentiable at x=c $\in$ (a, b) iff ${}_{x \to a}^{\lim}{\frac{f (x) - f (c)}{x - c}}$ or equivalently $\frac{f (c + h) - f (c)}{h}$ exist infinitely, and the value of this limit is called the derivative of f(x) at x=c and is denoted by f'(c)

A function f(x) is said to be

1. differentiable in an open interval (a,b) if it is derivable at every point of the interval (a, b)

2. A differentiable function if it is differentiable at every point of its domain.

If Y = f(x) is a differentiable function then the differentiable coefficient of 'y' w.r.t x denoted by $\frac{dy}{dx}$ = f'(x)

= ${}_{Δ x \to 0}^{\lim}{\frac{Δ y}{Δ x}}$ = ${}_{Δ x \to 0}^{\lim}{\frac{f (x - Δ x)}{Δ x}}$ where $Δ y$ is the increment in 'y' corresponding to a small increment $Δ x$ in x.

Differentiability and continuity

If a function f(x) is differentiable at a point 'a' in its domain then it is continuous at that point

Theorems on Differentiation.

If u, v, w are all differentiable functions of x then,

$\frac{d (c)}{dx}$ = 0 where 'c' is a constant.
$\frac{d}{dx}$ (u $\pm$ v)= $\frac{du}{dx}$ $\pm$ $\frac{dv}{dx}$
$\frac{d}{dx}$ (cu)=c. $\frac{du}{dx}$ where 'c' is a constant
$\frac{d}{dx}$ (uv)= $\frac{u . dv}{dx}$ + $\frac{v . du}{dx}$ (product rule)
$\frac{d}{dx}$ (uv w)= uv $\frac{dw}{dx}$ + uw. $\frac{dv}{dx}$ +v.w $\frac{du}{dx}$
$\frac{d}{dx}$ ( $\frac{u}{v}$ )= $\frac{v . \frac{du}{dx} - u . \frac{dv}{dx}}{v^{2}}$ (V $\neq$ 0 quotient rule)
$\frac{d}{dx}$ ( $\frac{1}{v}$ )= $\frac{- 1 dv}{v^{2} dx}$ (Reciprocal Rule)
$\frac{d}{dx}$ f(g(x)= f'[g(x)].g'(x) (function of a function rule)
If 'y' is a function of 'u' and 'u' is a function of 'x' then
$\frac{dy}{dx}$ = $\frac{dy}{du}$ x $\frac{du}{dx}$ (Chain rule)
If 'y' is a function of u, u is a function of v and v is a function of x then
$\frac{dy}{dx}$ = $\frac{dy}{du}$ x $\frac{du}{dv}$ x $\frac{dv}{dx}$ (Extension of chain rule)
$\frac{dy}{dx}$ = $\frac{1}{(dx / dy)}$ (inverse rule)

Implicit differentiation

If f 9x, y) =0 is an implicit function, then differentiate f (x, y)=0 term by term w. r.t to x and then solve for $\frac{dy}{dx}$ from resulting equation.

Parametric differentiation

if x= f (t) and y = g (t) then $\frac{dy}{dx}$ = $\frac{(dy / dt)}{(dx / dt)}$ = $\frac{g' (t)}{f' (t)}$ . In this case it is important to note that $\frac{d^{2} y}{{dx}^{2}}$ = $\frac{d}{dt} [\frac{dy}{dx}] x \frac{dt}{dx}$

Logarithmic Differentiation

If y = $u^{v}$ where u and v are any two functions of 'x' or 'y' then we take logarithm on both sides and differentiating ie $\frac{dy}{dx}$ = $\frac{d (u^{v})}{dx}$ = $u^{v} [\log u . \frac{dv}{dx} + v . \frac{d}{dx} (\log u)]$

Differentiation of a function w.r . to another functions of 'x' then $\frac{d [f (x)]}{d [g (x)]}$ = $\frac{f' (x)}{g' (x)}$

Results

1. $\frac{d}{dx} (x^{n}) = {nx}^{n - 1}$	2. $\frac{d}{dx} (\sqrt{x}$ )= $\frac{1}{\sqrt[2]{x}}$
3. $\frac{d}{dx} (e^{x}) = e^{x}$	4. $\frac{d}{dx}$ ( $a^{x}) = a^{x} . \log a$
5. $\frac{d}{dx}$ (\|x\|)= $\frac{x}{\| x \|}, x \neq 0$	6. $\frac{d}{dx} (\log x) = \frac{1}{x}$
7. $\frac{d}{dx} (\sin x) = \cos x$	8. $\frac{d}{dx} (\cos x) = - sinx$
9. $\frac{d}{dx}$ (tanx)= $\sec^{2} x$	10. $\frac{d}{dx} (cotx) = - {cosec}^{2} x$
11. $\frac{d}{dx} (\sec x) = Secx . tanx$	12. $\frac{d}{dx}$ (cosec x )=-cosec x.cot x
13. $\frac{d}{dx}$ ( $\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$	14. $\frac{d}{dx} (\cos^{- 1} x) = \frac{- 1}{\sqrt{1 - x^{2}}}$
15. $\frac{d}{dx} (\tan^{- 1} x) = \frac{1}{1 + x^{2}}$	16. $\frac{d}{dx} (\cot^{- 1} x) = \frac{- 1}{1 + x^{2}}$
17. $\frac{d}{dx} (\sec^{- 1} x) = \frac{1}{x \sqrt{x^{2} -} 1}$	18. $\frac{d}{dx} ({cosec}^{- 1} x) = \frac{- 1}{x . \sqrt{x^{2} - 1}}$
19. $\frac{d}{dx} (\sin hx) = \cos hx$	20. $\frac{d}{dx} (\cos hx) = \sin hx$
21. $\frac{d}{dx} (\tan hx) = \sec h^{2} x$	22. $\frac{d}{dx} (\cot hx) = cosec h^{2} x$
23. $\frac{d}{dx} (\sec hx) = \sec hx . \cot hx$	24. $\frac{d}{dx} (cosec hx) = cosec hx . \cot hx$
25. $\frac{d}{dx} (\sin h^{- 1} x) = \frac{1}{\sqrt{1 + x^{2}}}$	26. $\frac{d}{dx} (\cos h^{- 1} x) = \frac{1}{\sqrt{x^{2} - 1}}$
27. $\frac{d}{dx} (\tan h^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$	28. $\frac{d}{dx} (\cot h^{- 1} x) = \frac{1}{x^{2} - 1}$
29. $\frac{d}{dx} (\sec h^{- 1} x) = \frac{- 1}{x \sqrt{1 - x^{2}}}$	30. $\frac{d}{dx} (cosec h^{- 1} x) = \frac{1}{x . \sqrt{{1 + x}^{2}}}$

Successive Differentiation

If y = f(x), then $\frac{dy}{dx}$ is the first derivative of 'y' w.r. to x.

$\frac{d}{dx}$ ( $\frac{dy}{dx}$ )= $\frac{d^{2} y}{d x^{2}}$ = $y_{2}$ is the second derivative of 'y' w.r.to x and so on. In general differentiating 'y' successively 'n' times w.r.to 'x' we get $\frac{d}{dx}$ $(\frac{d^{n - 1} y}{{dx}^{n - 1}})$ = $\frac{d^{n} y}{{dx}^{n}}$ = $y_{n}$ is the $n^{th}$ derivative of 'y' w.r.to x.

Standard $n^{th}$ derivatives

If y = ${(ax + b)}^{n}$ then $y_{n}$ = $a^{n}$ .n!
If ylog (ax+b) then $y_{n}$ = $\frac{{(- 1)}^{n} n! a^{n}}{{(ax + b)}^{n}}$
If y = log (ax+b), then $y_{n}$ = $\frac{{(- 1)}^{n - 1} (n - 1)! a^{n}}{{(ax + b)}^{n}}$
If y = $e^{ax + b}$ , then $y_{n}$ = $a^{n} . e^{ax + b}$
If y = $a^{x}$ , then $y_{n}$ = ${(loga)}^{n}$ . $a^{x}$
If y = sin(ax+b) then $y_{n}$ = $a^{n}$ . sin(ax+b+ $\frac{n π}{2}$ )
If y= Cos (ax+b) then $y_{n}$ = $a^{n} \cos (ax + b + \frac{n π}{2})$