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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Differentiate each of the functions with respect to 'x' in Exercises 43 to 46 using first principle.

$c o s (x^{2} + 1)$

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t f (x) = c o s (x^{2} + 1) \dots (i) \\ \Rightarrow f (x + Δ x) = c o s [{(x + Δ x)}^{2} + 1] \dots (i i) \\ S u b t r a c t i n g e q n . (i) f r o m e q n . (i i) \\ f (x + Δ x) - f (x) = c o s [{(x + Δ x)}^{2} + 1] - c o s (x^{2} + 1) \\ D i v i d i n g b o t h s i d e s b y Δ x w e g e t \\ \frac{f (x + Δ x) - f (x)}{Δ x} = \frac{c o s [{(x + Δ x)}^{2} + 1] - c o s (x^{2} + 1)}{Δ x} \\ \underset{Δ x \to 0}{l i m} \frac{f (x + Δ x) - f (x)}{Δ x} = \underset{Δ x \to 0}{l i m} \frac{c o s [{(x + Δ x)}^{2} + 1] - c o s (x^{2} + 1)}{Δ x} \\ f' (x) = \underset{Δ x \to 0}{l i m} \frac{c o s [{(x + Δ x)}^{2} + 1] - c o s (x^{2} + 1)}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{- 2 s i n [\frac{{(x + Δ x)}^{2} + 1 + x^{2} + 1}{2}] . s i n [\frac{{(x + Δ x)}^{2} + 1 - x^{2} - 1}{2}]}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{- 2 s i n [\frac{x^{2} + Δ x^{2} + 2 x Δ x + x^{2} + 2}{2}] . s i n [\frac{x^{2} + Δ x^{2} + 2 x Δ x - x^{2}}{2}]}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{- 2 s i n [x^{2} + \frac{Δ x^{2}}{2} + x Δ x + 1] . s i n [Δ x \frac{(Δ x + 2 x)}{2}]}{Δ x} \\ = \underset{Δ x \to 0}{l i m} \frac{- 2 s i n [x^{2} + \frac{Δ x^{2}}{2} + x Δ x + 1] . s i n [Δ x \frac{(Δ x + 2 x)}{2}]}{Δ x [\frac{Δ x + 2 x}{2}]} * [\frac{Δ x + 2 x}{2}] \\ = \underset{Δ x [\frac{Δ x + 2 x}{2}] \to 0}{l i m} - 2 s i n [x^{2} + \frac{Δ x^{2}}{2} + x Δ x + 1] \frac{s i n [Δ x \frac{(Δ x + 2 x)}{2}]}{Δ x [\frac{Δ x + 2 x}{2}]} * [\frac{Δ x + 2 x}{2}] \\ T a k i n g limit, w e h a v e \\ = - 2 s i n (x^{2} + 1) . 1 . (x) = - 2 x s i n (x^{2} + 1) . \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\frac{1}{a x^{2} + b x + c}$

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{1}{a x^{2} + b x + c} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{1}{a x^{2} + b x + c}) \\ = \frac{(a x^{2} + b x + c) \frac{d}{d x} (1) - 1 . \frac{d}{d x} (a x^{2} + b x + c)}{{(a x^{2} + b x + c)}^{2}} [Using q u o t i e n t r u l e] \\ = \frac{(a x^{2} + b x + c) * 0 - (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} = \frac{- (2 a x + b)}{{(a x^{2} + b x + c)}^{2}} \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$s i n^{3} x c o s^{3} x$

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = s i n^{3} x c o s^{3} x \\ \frac{d y}{d x} = \frac{d}{d x} (s i n^{3} x c o s^{3} x) \\ = s i n^{3} x \frac{d}{d x} (c o s^{3} x) + c o s^{3} x . \frac{d}{d x} (s i n^{3} x) [Using p r o d u c t r u l e] \\ = s i n^{3} x . 3 c o s^{2} x (- s i n x) + c o s^{3} x . 3 s i n^{2} x . c o s x \\ = - 3 s i n^{4} x . c o s^{2} x + 3 c o s^{4} x . s i n^{2} x \\ = 3 s i n^{2} x . c o s^{2} x (- s i n^{2} x + c o s^{2} x) = 3 s i n^{2} x . c o s^{2} x . c o s 2 x \\ = \frac{3}{4} . 4 s i n^{2} x . c o s^{2} x . c o s 2 x = \frac{3}{4} {(2 s i n x . c o s x)}^{2} . c o s 2 x \\ = \frac{3}{4} s i n^{2} 2 x . c o s 2 x \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$x^{2} s i n x + c o s 2 x$

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = x^{2} s i n x + c o s 2 x \\ \frac{d y}{d x} = \frac{d}{d x} (x^{2} s i n x + c o s 2 x) \\ = \frac{d}{d x} (x^{2} s i n x) + \frac{d}{d x} (c o s 2 x) \\ = x^{2} c o s x + s i n x . 2 x + (- 2 s i n 2 x) \\ = x^{2} c o s x + 2 x s i n x - 2 s i n 2 x \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

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

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = {(2 x - 7)}^{2} {(3 x + 5)}^{3} \\ \frac{d y}{d x} = \frac{d}{d x} {(2 x - 7)}^{2} {(3 x + 5)}^{3} \\ = {(2 x - 7)}^{2} \frac{d}{d x} {(3 x + 5)}^{3} + {(3 x + 5)}^{3} \frac{d}{d x} {(2 x - 7)}^{2} [Using p r o d u c t r u l e] \\ = {(2 x - 7)}^{2} . 3 {(3 x + 5)}^{2} . 3 + {(3 x + 5)}^{3} . 2 (2 x - 7) . 2 \\ = 9 {(2 x - 7)}^{2} {(3 x + 5)}^{2} + 4 {(3 x + 5)}^{3} (2 x - 7) \\ = (2 x - 7) {(3 x + 5)}^{2} [9 (2 x - 7) + 4 (3 x + 5)] \\ = (2 x - 7) {(3 x + 5)}^{2} [18 x - 63 + 12 x + 20] \\ = (2 x - 7) {(3 x + 5)}^{2} (30 x - 43) = (2 x - 7) (30 x - 43) {(3 x + 5)}^{2} \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
${(s i n x + c o s x)}^{2}$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = {(s i n x + c o s x)}^{2} \\ \frac{d y}{d x} = \frac{d}{d x} {(s i n x + c o s x)}^{2} \\ = 2 (s i n x + c o s x) \frac{d}{d x} (s i n x + c o s x) \\ = 2 (s i n x + c o s x) (c o s x - s i n x) \\ = 2 (c o s^{2} x - s i n^{2} x) = 2 c o s 2 x [? c o s 2 x = c o s^{2} x - s i n^{2} x] \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

37. $\frac{a + b s i n x}{c + d c o s x}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{a + b s i n x}{c + d c o s x} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{a + b s i n x}{c + d c o s x}) \\ = \frac{(c + d c o s x) \frac{d}{d x} (a + b s i n x) - (a + b s i n x) \frac{d}{d x} (c + d c o s x)}{{(c + d c o s x)}^{2}} [Using q u o t i e n t r u l e] \\ = \frac{(c + d c o s x) (b c o s x) - (a + b s i n x) \frac{d}{d x} (- d s i n x)}{{(c + d c o s x)}^{2}} \\ = \frac{c b c o s x + b d c o s^{2} x + a d s i n x + b d s i n^{2} x}{{(c + d c o s x)}^{2}} \\ = \frac{c b c o s x + a d s i n x + b d (s i n^{2} x + c o s^{2} x)}{{(c + d c o s x)}^{2}} \\ = \frac{c b c o s x + a d s i n x + b d}{{(c + d c o s x)}^{2}} \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$(a x^{2} + c o t x) (p + q c o s x)$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = (a x^{2} + c o t x) (p + q c o s x) \\ \frac{d y}{d x} = \frac{d}{d x} (a x^{2} + c o t x) (p + q c o s x) \\ = (a x^{2} + c o t x) \frac{d}{d x} (p + q c o s x) + (p + q c o s x) \frac{d}{d x} (a x^{2} + c o t x) [Using p r o d u c t r u l e] \\ = (a x^{2} + c o t x) (- q s i n x) + (p + q c o s x) (2 a x - c o s e c^{2} x) \\ = - q s i n x (a x^{2} + c o t x) + (p + q c o s x) (2 a x - c o s e c^{2} x) \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\frac{x^{2} c o s \frac{π}{4}}{s i n x}$

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{x^{2} c o s \frac{π}{4}}{s i n x} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{x^{2} c o s \frac{π}{4}}{s i n x}) \\ = c o s \frac{π}{4} . \frac{d}{d x} (\frac{x^{2}}{s i n x}) \\ = \frac{\frac{1}{\sqrt[]{2}} [s i n x \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} (s i n x)]}{s i n^{2} x} [q u o t i e n t r u l e] \\ = \frac{1}{\sqrt[]{2}} . [\frac{s i n x . 2 x - x^{2} c o s x}{s i n^{2} x}] = \frac{1}{\sqrt[]{2}} [\frac{2 x}{s i n x} - \frac{x^{2} c o s x}{s i n^{2} x}] \\ = \frac{1}{\sqrt[]{2}} [2 x c o s e c x - x^{2} c o t x c o s e c x] = \frac{x}{\sqrt[]{2}} c o s e c x [2 - x c o t x] \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$\frac{x^{5} - c o s x}{s i n x}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t y = \frac{x^{5} - c o s x}{s i n x} \\ \frac{d y}{d x} = \frac{d}{d x} (\frac{x^{5} - c o s x}{s i n x}) \\ = \frac{s i n x \frac{d}{d x} (x^{5} - c o s x) - (x^{5} - c o s x) \frac{d}{d x} (s i n x)}{s i n^{2} x} [q u o t i e n t r u l e] \\ = \frac{s i n x (5 x^{4} + s i n x) - (x^{5} - c o s x) (c o s x)}{s i n^{2} x} \\ = \frac{5 x^{4} . s i n x + s i n^{2} x - x^{5} c o s x + c o s^{2} x}{s i n^{2} x} \\ = \frac{5 x^{4} . s i n x - x^{5} c o s x + (s i n^{2} x + c o s^{2} x)}{s i n^{2} x} = \frac{5 x^{4} . s i n x - x^{5} c o s x + 1}{s i n^{2} x} \end{array}$

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