Successive Differentiation | Btech Shots!

Successive Differentiation

Consider a function. We differentiate it one time, that is called first derivative. We differentiate it second time, that is called second derivative. We differentiate it third time, that is called third derivative. We differentiate it nth time, that is called nth derivative. This whole process is called Successive Differentiation.
nth derivative of some standard functions:
$1) \frac{d^{n}}{d x^{n}} (c o n s t a n t) = 0$ $2) \frac{d^{n}}{d x^{n}} (x^{n}) = n!$ $3) \frac{d^{n}}{d x^{n}} (x^{m}) = \frac{m!}{(m - n)!} x^{m - n}, m \neq n$ $4) \frac{d^{n}}{d x^{n}} (a x + b)^{n} = n! (a)^{n}$ $5) \frac{d^{n}}{d x^{n}} (a x + b)^{m} = \frac{m!}{(m - n)!} a^{n} (a x + b)^{m - n}$ $6) \frac{d^{n}}{d x^{n}} (e^{a x}) = a^{n} e^{a x}$ $7) \frac{d^{n}}{d x^{n}} (e^{a x + b}) = a^{n} e^{a x + b}$ $8) \frac{d^{n}}{d x^{n}} (a^{m x}) = m^{n} a^{m x} (l o g a)^{n}$ $9) \frac{d^{n}}{d x^{n}} [l o g (a x + b)] = \frac{(- 1)^{n} (n - 1)! a^{n}}{(a x + b)^{n}}$ $10) \frac{d^{n}}{d x^{n}} [\sin a x + b] = a^{n} \sin [\frac{n π}{2} + (a x + b)]$ $11) \frac{d^{n}}{d x^{n}} [\cos a x + b] = a^{n} \cos [\frac{n π}{2} + (a x + b)]$ $12) \frac{d^{n}}{d x^{n}} (\frac{1}{a x + b}) = \frac{(- 1)^{n} n! a^{n}}{(a x + b)^{n + 1}}$ $13) \frac{d^{n}}{d x^{n}} [\frac{1}{(a x + b)^{2}}] = \frac{(- 1) (n + 1)! a^{n}}{(a x + b)^{n + 2}}$ $14) \frac{d^{n}}{d x^{n}} [e^{a x} \sin (b x + c)] = e^{a x} (a^{2} + b^{2})^{n / 2} \sin [b x + c + n \tan^{- 1} (\frac{b}{a})]$ $15) \frac{d^{n}}{d x^{n}} [e^{a x} \cos (b x + c)] = e^{a x} (a^{2} + b^{2})^{n / 2} \cos [b x + c + n \tan^{- 1} (\frac{b}{a})]$ $16) \frac{d^{n}}{d x^{n}} [f (x) \pm g (x)] = \frac{d^{n}}{d x^{n}} [f (x)] \pm \frac{d^{n}}{d x^{n}} [g (x)]$ In successive differentiation what we do is we find first 3 or 4 derivatives, find a pattern and generalise a formula for nth derivative. Following examples will help you understand better.

Sample Problem 1

Find the nth derivative of $x^{n}$ .

Solution

Let $y = x^{n}$
First derivative: $y_{1} = n x^{n - 1}$ Second derivative: $y_{2} = (n) (n - 1) x^{n - 2}$ Third derivative: $y_{3} = (n) (n - 1) (n - 2) x^{n - 3}$ Fourth derivative: $y_{4} = (n) (n - 1) (n - 2) (n - 3) x^{n - 4}$ Similarly, nth derivative $y_{n} = (n) (n - 1) (n - 2) (n - 3) . . . . . .1 . x^{n - n}$ $\Rightarrow y_{n} = n!$

Sample Problem 2

Find the nth derivative of $\sin (a x + b)$

Solution

$y = \sin (a x + b)$ $y_{1} = \cos (a x + b) (a)$ $= a \cos (a x + b)$ $= a \sin [\frac{π}{2} + (a x + b)]$ $y_{2} = a [\cos (\frac{π}{2} + (a x + b))] (a)$ $= a^{2} \sin [\frac{π}{2} + \frac{π}{2} + (a x + b)]$ $= a^{2} \sin [\frac{2 π}{2} + (a x + b)]$ $y_{3} = a^{2} \cos [\frac{2 π}{2} + (a x + b)] (a)$ $= a^{3} \sin [\frac{3 π}{2} + (a x + b)]$ Similarly, $y_{n} = a^{n} \sin [\frac{n π}{2} + (a x + b)]$

Sample Problem 3

Find the nth derivative of $l o g (4 x^{2} - 1)$

Solution

We don't have formula for quadratic term inside log but we have formula for linear term inside log. So first we will factorise the quadratic term into linear term and then apply the formula. $y = \log (4 x^{2} - 1)$ $= \log (2 x + 1) (2 x - 1)$ $y = \log (2 x + 1) + \log (2 x - 1)$ As we know, $\frac{d^{n}}{d x^{n}} [l o g (a x + b)] = \frac{(- 1)^{n} (n - 1)! a^{n}}{(a x + b)^{n}}$ $y_{n} = \frac{(- 1)^{n - 1} (n - 1)! (2)^{n}}{(2 x + 1)^{n}} + \frac{(- 1)^{n - 1} (n - 1)! (2)^{n}}{(2 x - 1)^{n}}$ $= (- 1)^{n - 1} (n - 1)! (2)^{n} [\frac{1}{(2 x + 1)^{n}} + \frac{1}{(2 x - 1)^{n}}]$

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