Chapter 13: Limits and Derivatives

Ans: Given:

$(3x + 5)(1 + \tan x)$

We know that,

Differentiate the given equation w.r.t \[x\] using product rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

Then,$\quad y = (3x + 5)(1 + \tan x)$

Differentiate on both side w.r.t \[x\]

$\therefore \quad \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}[(3x + 5)(1 + \tan x)]$

$= (3x + 5)\dfrac{d}{{dx}}(1 + \tan x) + (1 + \tan x)\dfrac{d}{{dx}}(3x + 5){\text{ }}\left[ {{\text{product rule}}} \right]$

$= (3x + 5)\left( {{{\sec }^2}x} \right) + (1 + \tan x) \cdot 3$

$= (3x + 5){\sec ^2}x + 3(1 + \tan x)$

$= 3x{\sec ^2}x + 5{\sec ^2}x + 3\tan x + 3$

Ans: Given:

$(\sec x - 1)(\sec x + 1)$

We know that, Differentiate the given equation w.r.t \[x\] using chain rule

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

$\because (a + b)(a - b) = {a^2} - {b^2}$

Let, $y = (\sec x - 1)(\sec x + 1)$

$y = \left( {{{\sec }^2} - 1} \right)$$\left[ {\because (a + b)(a - b) = {a^2} - {b^2}} \right]$

$= {\tan ^2}x$

$\therefore \quad \dfrac{{dy}}{{dx}} = 2\tan x \cdot \dfrac{d}{{dx}}\tan x$

$= 2\tan x \cdot {\sec ^2}x$                    (By chain rule)

Ans: Given:

$\dfrac{{3x + 4}}{{5{x^2} - 7x + 9}}$

We know that,

Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}\]

Let,  $\quad y = \dfrac{{3x + 4}}{{5{x^2} - 7x + 9}}$

$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{\left( {5{x^2} - 7x + 9} \right)\dfrac{d}{{dx}}(3x + 4) - (3x + 4)\dfrac{d}{{dx}}\left( {5{x^2} - 7x + 9} \right)}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}{\text{ }}\left[ {{\text{by quotient rule}}} \right]$

${\text{       }} = \dfrac{{\left( {5{x^2} - 7x + 9} \right) \cdot 3 - (3x + 4)(10x - 7)}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}$

${\text{       }} = \dfrac{{15{x^2} - 21x + 27 - 30{x^2} + 21x - 40x + 28}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}$

${\text{       }} = \dfrac{{ - 15{x^2} - 40x + 55}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}$

${\text{       }} = \dfrac{{55 - 15{x^2} - 40x}}{{{{\left( {5{x^2} - 7x + 9} \right)}^2}}}$

Ans: Given:

$\dfrac{{{x^5} - \cos x}}{{\sin x}}$

We know that, Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}\]

Then,Let $y = \dfrac{{{x^5} - \cos x}}{{\sin x}}$

$\dfrac{{dy}}{{dx}} = \dfrac{{\sin x\dfrac{d}{{dx}}\left( {{x^5} - \cos x} \right) - \left( {{x^5} - \cos x} \right)\dfrac{{dy}}{{dx}}\sin x}}{{{{\left( {\sin x} \right)}^2}}}$    (by quotient rule)

$= \dfrac{{\sin x\left( {5{x^4} + \sin x} \right) - \left( {{x^5} - \cos x} \right)\cos x}}{{{{\sin }^2}x}}$

$= \dfrac{{5{x^4}\sin x + {{\sin }^2}x - {x^5}\cos x + {{\cos }^2}x}}{{{{\sin }^2}x}}$

$= \dfrac{{5{x^4}\sin x - {x^5}\cos x + {{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}}$

$= \dfrac{{5{x^4}\sin x - {x^5}\cos x + 1}}{{{{\sin }^2}x}}$

Ans: Given:

\[\dfrac{{{x^2}\cos \dfrac{\pi }{4}}}{{\sin x}}\]

We know that, Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}\]

Let \[y = \dfrac{{{x^2}\cos \dfrac{\pi }{4}}}{{\sin x}}\]

\[y = \dfrac{1}{{\sqrt 2 }} \cdot \dfrac{{{x^2}}}{{\sin x}}\]

$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt 2 }}\left[ {\dfrac{{\sin x \cdot \dfrac{d}{{dx}}{x^2} - {x^2}\dfrac{d}{{dx}}\sin x}}{{{{\sin }^2}x}}} \right]$       (by quotient rule)

$= \dfrac{1}{{\sqrt 2 }}\left[ {\dfrac{{\sin x \cdot 2x - {x^2} \cdot \cos x}}{{{{\sin }^2}x}}} \right]$

$= \dfrac{1}{{\sqrt 2 }} \cdot \dfrac{{2x\sin x - {x^2}\cos x}}{{{{\sin }^2}x}}$

$= \dfrac{x}{{\sqrt 2 }}[2\operatorname{cosec} x - x\cot x\operatorname{cosec} x]$

$= \dfrac{x}{{\sqrt 2 }}\operatorname{cosec} [2 - x\cot x]$

Ans: Given:

\[\left( {a{x^2} + \cot x} \right)\left( {p + q\cos x} \right)\]

We know that, Differentiate the given equation w.r.t \[x\] using product rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u\]

Let \[y = \left( {a{x^2} + \cot x} \right)\left( {p + q\cos x} \right)\]

$\therefore \quad \dfrac{{dy}}{{dx}} = \left( {a{x^2} + \cot x} \right)\dfrac{d}{{dx}}(p + q\cos x) + (p + q\cos x)\dfrac{d}{{dx}}\left( {a{x^2} + \cot x} \right){\text{ }}\left[ {{\text{by product rule}}} \right]$

$= \left( {a{x^2} + \cot x} \right)( - q\sin x) + (p + q\cos x)\left( {2ax - {{\operatorname{cosec} }^2}x} \right)$

$=  - q\sin x\left( {a{x^2} + \cot x} \right) + (p + q\cos x)\left( {2ax - {{\operatorname{cosec} }^2}x} \right)$

Ans: Given:

$\dfrac{{a + b\sin x}}{{c + d\cos x}}$

We know that, Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}\]

Let, $y = \dfrac{{a + b\sin x}}{{c + d\cos x}}$

$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{(c + d\cos x)\dfrac{d}{{dx}}(a + b\sin x) - (a + b\sin x)\dfrac{d}{{dx}}(c + d\cos x)}}{{{{(c + d\cos x)}^2}}}{\text{ }}\left[ {{\text{by quotient rule}}} \right]$

$= \dfrac{{(c + d\cos x)(b\cos x) - (a + b\sin x)( - d\sin x)}}{{{{(c + d\cos x)}^2}}}$

$= \dfrac{{bc\cos x + bd{{\cos }^2}x + ad\sin x + bd{{\sin }^2}x}}{{{{(c + d\cos x)}^2}}}$

$= \dfrac{{bc\cos x + ad\sin x + bd\left( {{{\cos }^2}x + {{\sin }^2}x} \right)}}{{{{(c + d\cos x)}^2}}}$

$= \dfrac{{bc\cos x + ad\sin x + bd}}{{{{(c + d\cos x)}^2}}}$

Ans: Given:

${\left( {\sin x + \cos x} \right)^2}$

We know that, Differentiate the given equation w.r.t \[x\] using quotient rule

\[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

\[\because \cos 2x = {\cos ^2}x - {\sin ^2}x\]

\[{\text{Let }}y = {\left( {\sin x + \cos x} \right)^2}\]

\[\therefore \dfrac{{dy}}{{dx}} = 2\left( {\sin x + \cos x} \right)\left( {\sin x - \cos x} \right)\](By chain rule)

\[{\text{       }} = 2\left( {{{\cos }^2}x - {{\sin }^2}x} \right){\text{                   }}\left[ {\because \cos 2x = {{\cos }^2}x - {{\sin }^2}x} \right]\]

\[{\text{       }} = 2\cos 2x\]

Ans: Given:

${(2x - 7)^2}{(3x + 5)^3}$

We know that, Differentiate the given equation w.r.t \[x\] using product rule and chain rule

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\] and \[\because \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]

Let $y = {(2x - 7)^2}{(3x + 5)^3}$

$\therefore \dfrac{{dy}}{{dx}} = {(2x - 7)^2}\dfrac{d}{{dx}}{(3x + 5)^3} + {(3x + 5)^3}\dfrac{d}{{dx}}{(2x - 7)^2}{\text{   }}\left[ {{\text{by product rule}}} \right]$

$= {(2x - 7)^2}(3){(3x + 5)^2}(3) + {(3x + 5)^3}2(2x - 7)(2){\text{   }}\left[ {{\text{by chain rule}}} \right]$

$= 9{(2x - 7)^2}{(3x + 5)^2} + 4{(3x + 5)^3}(2x - 7)$

$= (2x - 7){(3x + 5)^2}[9(2x - 7) + 4(3x + 5)]$

$= (2x - 7){(3x + 5)^2}(18x - 63 + 12x + 20)$

$= (2x - 7){(3x + 5)^2}(30x - 43)$

Ans: Given:

${x^2}\sin x + \cos 2x$

We know that, Differentiate the given equation w.r.t \[x\] using product rule 

\[\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}u + u\dfrac{d}{{dx}}v\]

${\text{Let }}y = {x^2}\sin x + \cos 2x$

$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {{x^2}\sin x} \right)}}{{dx}} + \dfrac{{d\left( {\cos 2x} \right)}}{{dx}}$

${\text{       }} = \left( {2x\sin x + {x^2} \cdot \cos x} \right) + \left( { - \sin 2x \cdot 2} \right){\text{ }}\left[ {{\text{By product rule}}} \right]$

${\text{       }} = 2x\sin x + {x^2} \cdot \cos x - 2\sin 2x\left[ {{\text{By chain rule}}} \right]{\text{       }}$