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New answer posted

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Maths Limits and Derivatives

47. (ax + b) (cx + d)²

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alok kumar singh

Contributor-Level 10

47. Given, f (x) = (ax + b) (cx + d)²

So, f?(x) = (ax +b) $\frac{d}{d x} {(c x + d)}^{2} + {(c x + d)}^{2} \frac{d}{d x} (a x + b)$

$= (a x + b) \frac{d}{d x} [(c + d) (c x + d)] + {(c x + d)}^{2} a$

$= (a x + b) [(c + d) \frac{d}{d x} (x + d) + (c + d) \frac{d}{d x} (c + d)] + a {(c x + d)}^{2}$

$= (a x + b) [c (c x + d) + c (c x + d)] + a {(c x + d)}^{2}$

$= (a x + b) \cdot 2 \cdot c (c x + d) + a {(c x + d)}^{2}$

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New answer posted

a year ago

Maths Limits and Derivatives

44. Find the derivative of the following functions from first principle:

(i) x (ii) (x)1 (iii) sin(x + 1) (iv) cos $(x - \frac{π}{8})$

Find the derivative of the following functions (it is to be understood that a, b, c, d,

p, q, r and s are fixed non-zero constants and m and n are integers):

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alok kumar singh

Contributor-Level 10

44.

= \underset{h \to 0}{l i m} \frac{- x - h + x}{h}

= \underset{h \to 0}{l i m} \frac{- h}{h}

= \underset{h \to 0}{l i m} - 1

= - 1.

(ii) Given, f(x) = (-x)^-¹

by first principle,

f(x) $\underset{h \to 0}{l i m} \frac{[-] {(x + h)}^{- 1} - {(- x)}^{- 1}}{h}$

(iii) Given, f(x) = sin(x + 1)

By first principle,

f'(x) = $\underset{h \to 0}{l i m} \frac{f (x + h) - f (x)}{h}$

$= \underset{h \to 0}{l i m} \frac{s i n (x + h + 1) - s i n (x + 1)}{h}$

$= \underset{h \to 0}{l i m} \frac{2}{h} \cdot c o s (\frac{x + h + 1 + x + 1}{2}) \cdot s i n (\frac{x + h + 1 - (x + 1)}{2})$

$= \underset{h \to 0}{l i m} \frac{2}{h} c o s (\frac{2 x + 2 + h}{2}) s i n (\frac{h}{2}) .$

$= \underset{h \to 0}{l i m} c o s (\frac{2 x + 2 + h}{2}) \underset{h \to 0}{l i m} \frac{s i n \frac{h}{2}}{\frac{h}{2}}$

$= c o s (\frac{2 x + 2 + 0}{2}) * 1 .$

= cos (x + 1)

(iv) Given, f(x) = cos $(x - \frac{π}{8})$

By first principle,

f(x) = $\underset{h \to 0}{l i m} \frac{f (x + h) - f (x)}{h}$

$= \underset{h \to 0}{l i m} \frac{c o s (x + h - \frac{π}{8}) - c o s (x - \frac{π}{8})}{h}$

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New answer posted

a year ago

Maths Limits and Derivatives

43. Find the derivative of the following functions:

(i) sin x cos x (ii) $s e c x$ (iii) 5 sec x + 4 cosx

(iv) $c o s e c x$ (v) 3cot x + 5 cosec x

(vi) $5 s i n x - 6 c o s x + 7$ (vii) $2 t a n x - 7 s e c x$

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alok kumar singh

Contributor-Level 10

. (i) f(x)=sin x cos x

So, $f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$

$= \underset{h ? 0}{l i m} \frac{s i n (x + h) c o s (x + h) ? s i n x c o s x}{h}$

$= \underset{h ? 0}{l i m} \frac{1}{2 h} * [2 s i n (x + h) c o s (x + h) ? 2 s i n x c o s x]$

$= \underset{h ? 0}{l i m} \frac{1}{2 h} [s i n 2 (x + h) ? s i n 2 x]$

$= \underset{h ? 0}{l i m} \frac{1}{2 h} [2 c o s \frac{2 (x + h) + 2 x}{2} s i n \frac{2 (x + h) ? 2 x}{2}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [c o s (2 x + h) s i n h]$

$= \underset{h ? 0}{l i m} c o s (2 x + h) * \underset{h ? 0}{l i m} \frac{s i n h}{h}$

$= c o s (2 x + 0)$

$= c o s 2 x$

$(ii) f (x) = s e c x$

So, $f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$

$= \underset{h ? 0}{l i m} \frac{1}{h} [s e c (x + h) ? s e c x]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{1}{c o s (x + h)} ? \frac{1}{c o s x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{c o s x ? c o s (x + h)}{c o s (x + h) c o s x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{? 2 s i n (\frac{x + x + h}{2}) s i n (\frac{x ? (x + h)}{2})}{c o s (x + h) c o s x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [? \frac{2 s i n (\frac{2 x + h}{2}) s i n (? h / 2)}{c o s (x + h) c o s x}]$

$= \underset{h ? 0}{l i m} \frac{(? 1 s i n (\frac{2 x + h}{2})}{c o s (x + h) c o s x} * \underset{h ? 0}{l i m} (? 1) \frac{s i n h / 2}{h / 2}$

$= \frac{s i n x}{c o s x ? c o s x} * 1$

$= t a n x ? s e c x .$

(iii) Given f(x)=5 sec x+4 cosx.

So, $f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$

$\begin{array}{l} = \underset{h ? 0}{l i m} \frac{5 s e c (x + h) + 4 c o s (x + h) ? [5 s e c x + 4 c o s x]}{h} . \end{array}$

$= \underset{h ? 0}{l i m} \frac{5}{h} [s e c (x + h) ? s e c x] + \underset{h ? 0}{l i m} \frac{4}{h} [c o s (x + h) ? c o s x]$

$= \underset{h ? 0}{l i m} \frac{5}{h} [\frac{1}{c o s (x + h)} ? \frac{1}{c o s x}] + \underset{h ? 0}{l i m} \frac{4}{h} [? 2 s i n (\frac{x + h + x}{2}) s i n (\frac{x + h ? x}{2})]$

$= \underset{h ? 0}{l i m} \frac{5}{h} [\frac{c o s x ? c o s (x + h)}{c o s (x + h) (c o s x)}] + \underset{h ? 0}{l i m} \frac{4}{h} [? 2 s i n (\frac{2 x + h}{2}) s i n \frac{h}{2}]$

$= \underset{h ? 0}{l i m} \frac{5}{h} [? \frac{2 s i n (\frac{2 x + h}{2}) s i n (? h / 2)}{c o s (x + h) c o s x}] ? 4 \underset{h ? 0}{l i m} s i n (\frac{2 x + h}{2}) \underset{h ? 0}{l i m} \frac{s i n h / 2}{h / 2}$

$= \frac{s i n (\frac{2 x + 0}{2})}{c o s (x + 0) c o s x} * 1 ? 4 s i n (\frac{2 x}{2})$

$= 5 \frac{s i n x}{c o s x} ? \frac{1}{c o s x} ? 4 s i n x$

$= 5 t a n x ? s e c x ? 4 s i n x$

$(iv) Given f (x) = c o s e c x$

$f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$

$= \underset{h ? 0}{l i m} \frac{1}{h} [c o s e c (x + h) ? c o s e c x]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{1}{s i n (x + h)} ? \frac{1}{s i n x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{s i n x ? s i n (x + h)}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{2 c o s (\frac{x + x + h}{2}) s i n (\frac{x ? (x + h)}{2})]}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{2 c o s (\frac{x + x + h}{2}) s i n (\frac{x ? (x + h)}{2})}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{1}{h} [\frac{2 c o s (\frac{2 x + h}{2}) s i n (? h / 2)]}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{c o s (\frac{2 x + h}{2})}{s i n (x + h) s i n x} * \frac{(? 1) s i n (2)}{h / 2})$

$= \frac{c o s (\frac{2 x + 0}{2})}{s i n (x + 0) s i n x} * (? 1)$

$= ? \frac{c o s x}{s i n x} * \frac{1}{s i n x}$

$= ? c o t x ? c o s e c x$

(v) Given,f(x)=3 cot x+5cosecx.

So, $f^{?} (x) = \underset{h ? 0}{l i m} \frac{f (x + h) ? f (x)}{h}$ $= \frac{2}{c o s (x + 0) c o s x} * 1 + 7 s i n (\frac{2 x + 0}{2}) * (? 1)$

$= \underset{h ? 0}{l i m} 3 c o t (x + h) + 5 c o s e c (x + h) ? [3 c o t x + 5 c o s x)$

$_{h ? 0} \frac{3}{h} [c o t (x + h) ? c o t x] + \underset{h ? 0}{l i m}$

$= ? \frac{5}{h} [c o s e c (x + h) ? c o s e c x]$

$= \underset{h ? 0}{l i m} \frac{3}{h} [\frac{c o s (x + h)}{s i n (x + h)} ? \frac{c o s x}{s i n x}] + \underset{h ? 0}{l i m} \frac{5}{h} [\frac{1}{s i n (x + h)} ? \frac{1}{s i n x}]$

$= \underset{h ? 0}{l i m} \frac{3}{h} [\frac{c o s (x + h) s i n x ? c o s x s i n (x + h)}{s i n (x + h) s i n x}] + \underset{h ? 0}{l i m} \frac{5}{h} [\frac{s i n x ? s i n (x + h)}{s i n (x + h) s i n x}]$

$= \underset{h ? 0}{l i m} \frac{3}{h} [\frac{s i n (x ? (x + h))}{s i n (x + h) s i n x} + \underset{h ? 0}{l i m} \frac{5}{h} [\frac{2 c o s (\frac{x + x + h}{2}) s i n (\frac{x ? (x + h)}{2})}{s i n (x + h) s i n x}$

$= \underset{h ? 0}{l i m} \frac{3}{s i n (x + h) s i n x} * \underset{h ? 0}{l i m} (? 1) \frac{s i n h}{h} + \underset{h ? 0}{l i m} \frac{5}{h} [\frac{2 c o s (\frac{2 x + h}{2}) s i n (? h / 2)}{s i n (x + h) s i n x}$

$= \frac{3}{s i n x ? s i n x} * (? 1) + 5 l i m$

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New answer posted a year ago

Maths Limits and Derivatives

43. Find the derivative of the following functions: (i) sin x cos x

(ii) $s e c x$ (iii) 5 sec x + 4 cosx

(iv) $c o s e c x$ (v) 3cot x + 5 cosec x

(vi) $5 s i n x - 6 c o s x + 7$ (vii) $2 t a n x - 7 s e c x$

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alok kumar singh

Contributor-Level 10

(i) f(x)=sin x cos x