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NCERT Exemplar · Class 11 Mathematics Limits and Derivatives

80 questions · 80 still being checked

EXERCISE 13.3 31–40 (part 4 of 8)

  1. Differentiate each of the functions w. r. to \(\displaystyle x\) in Exercises $\displaystyle 29$ to 42.

    Exercise 31

    (3x+5)(1+tan⁡x)\displaystyle (3 x+5)(1+\tan x)

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    NCERT’s answer
    \(\displaystyle 3 x \sec ^2 x+5 \sec ^2 x+3 \tan x+3\)
    Product rule with \(\displaystyle u=3x+5,\ v=1+\tan x\).\[\frac{dy}{dx}=u'v+uv' \]\[u'=3,\qquad v'=\sec^2 x \]\[\frac{dy}{dx}=3(1+\tan x)+(3x+5)\sec^2 x \]Answer: \(\displaystyle 3(1+\tan x)+(3x+5)\sec^2 x\)

  2. Exercise 32

    (sec⁡x−1)(sec⁡x+1)\displaystyle (\sec x-1)(\sec x+1)

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    NCERT’s answer
    \(\displaystyle 2 \tan x \sec ^2 x\)
    Simplify first.\[y=(\sec x-1)(\sec x+1)=\sec^2 x-1 \]Product rule on \(\displaystyle \sec x\cdot\sec x\), with \(\displaystyle \dfrac{d}{dx}\sec x=\sec x\tan x\):\[\frac{dy}{dx}=\sec x\cdot\sec x\tan x+\sec x\tan x\cdot\sec x-0 \]\[\frac{dy}{dx}=2\sec^2 x\tan x \]Answer: \(\displaystyle 2\sec^2 x\tan x\)

  3. Exercise 33

    3x+45x2−7x+9\displaystyle \frac{3 x+4}{5 x^2-7 x+9}

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    NCERT’s answer
    \(\displaystyle \frac{55-40 x-15 x^2}{\left(5 x^2-7 x+9\right)^2}\)
    Quotient rule: \[\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2} \] \[u=3x+4,\; u'=3,\qquad v=5x^2-7x+9,\; v'=10x-7 \] \[\frac{dy}{dx}=\frac{3(5x^2-7x+9)-(3x+4)(10x-7)}{(5x^2-7x+9)^2} \] \[=\frac{15x^2-21x+27-(30x^2+19x-28)}{(5x^2-7x+9)^2} \] \[=\frac{-15x^2-40x+55}{(5x^2-7x+9)^2} \] Answer: \(\displaystyle \dfrac{-15x^2-40x+55}{(5x^2-7x+9)^2}\)

  4. Exercise 34

    x5−cos⁡xsin⁡x\displaystyle \frac{x^5-\cos x}{\sin x}

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Quotient rule: \[u=x^5-\cos x,\; u'=5x^4+\sin x,\qquad v=\sin x,\; v'=\cos x \] \[\frac{dy}{dx}=\frac{(5x^4+\sin x)\sin x-(x^5-\cos x)\cos x}{\sin^2x} \] \[=\frac{5x^4\sin x+\sin^2x-x^5\cos x+\cos^2x}{\sin^2x} \] \[\sin^2x+\cos^2x=1 \] Answer: \(\displaystyle \dfrac{5x^4\sin x-x^5\cos x+1}{\sin^2x}\)

  5. Exercise 35

    x2cos⁡π4sin⁡x\displaystyle \frac{x^2 \cos \dfrac{\pi}{4}}{\sin x}

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    NCERT’s answer
    \(\displaystyle \frac{x}{\sqrt{2}} \operatorname{cosec} x(2-x \cot x)\)
    \(\displaystyle \cos\frac{\pi}{4}\) is a constant: \[\cos\frac{\pi}{4}=\frac{1}{\sqrt2},\qquad y=\frac{1}{\sqrt2}\cdot\frac{x^2}{\sin x} \] Quotient rule: \[\frac{dy}{dx}=\frac{1}{\sqrt2}\cdot\frac{2x\sin x-x^2\cos x}{\sin^2x} \] Answer: \(\displaystyle \dfrac{x\,(2\sin x-x\cos x)}{\sqrt2\,\sin^2x}\)

  6. Exercise 36

    (ax2+cot⁡x)(p+qcos⁡x)\displaystyle \left(a x^2+\cot x\right)(p+q \cos x)

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    NCERT’s answer
    \(\displaystyle \left(a x^2+\cot x\right)(-q \sin x)+(p+q \cos x)\left(2 a x-\operatorname{cosec}^2 x\right)\)
    Product rule: \[(uv)'=u'v+uv' \] \[u=ax^2+\cot x,\; u'=2ax-\operatorname{cosec}^2x \] \[v=p+q\cos x,\; v'=-q\sin x \] \[\frac{dy}{dx}=(2ax-\operatorname{cosec}^2x)(p+q\cos x)-q\sin x\,(ax^2+\cot x) \] \[q\sin x\cot x=q\cos x \] Answer: \(\displaystyle (2ax-\operatorname{cosec}^2x)(p+q\cos x)-aqx^2\sin x-q\cos x\)

  7. Exercise 37

    a+bsin⁡xc+dcos⁡x\displaystyle \frac{a+b \sin x}{c+d \cos x}

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{b c \cos x+a d \sin x+d b}{(c+d \cos x)^2}\)
    Quotient rule: \[u=a+b\sin x,\; u'=b\cos x,\qquad v=c+d\cos x,\; v'=-d\sin x \] \[\frac{dy}{dx}=\frac{b\cos x\,(c+d\cos x)+d\sin x\,(a+b\sin x)}{(c+d\cos x)^2} \] \[=\frac{bc\cos x+bd\cos^2x+ad\sin x+bd\sin^2x}{(c+d\cos x)^2} \] \[\cos^2x+\sin^2x=1 \] Answer: \(\displaystyle \dfrac{bc\cos x+ad\sin x+bd}{(c+d\cos x)^2}\)

  8. Exercise 38

    (sin⁡x+cos⁡x)2\displaystyle (\sin x+\cos x)^2

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    NCERT’s answer
    \(\displaystyle 2 \cos 2 x\)
    Expand first: \[y=\sin^2x+\cos^2x+2\sin x\cos x=1+2\sin x\cos x \] Product rule on \(\displaystyle \sin x\cos x\): \[\frac{dy}{dx}=2\left(\cos x\cdot\cos x+\sin x\cdot(-\sin x)\right) \] \[=2(\cos^2x-\sin^2x)=2\cos 2x \] Answer: \(\displaystyle 2\cos 2x\)

  9. Exercise 39

    (2x−7)2(3x+5)3\displaystyle (2 x-7)^2(3 x+5)^3

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle (2 x-7)(30 x-43)(3 x+5)^2\)
    Product rule, with \(\displaystyle \frac{d}{dx}(ax+b)^n=na(ax+b)^{n-1}\): \[\frac{d}{dx}(2x-7)^2=4(2x-7),\qquad \frac{d}{dx}(3x+5)^3=9(3x+5)^2 \] \[\frac{dy}{dx}=4(2x-7)(3x+5)^3+9(2x-7)^2(3x+5)^2 \] \[=(2x-7)(3x+5)^2\left[4(3x+5)+9(2x-7)\right] \] \[=(2x-7)(3x+5)^2(30x-43) \] Answer: \(\displaystyle (2x-7)(3x+5)^2(30x-43)\)

  10. Exercise 40

    x2sin⁡x+cos⁡2x\displaystyle x^2 \sin x+\cos 2 x

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle x^2 \cos x+2 x \sin x-2 \sin 2 x\)
    Product rule on the first term; for the second: \[\cos 2x=1-2\sin^2x \] \[\frac{d}{dx}\left(x^2\sin x\right)=2x\sin x+x^2\cos x \] \[\frac{d}{dx}\cos 2x=-4\sin x\cos x=-2\sin 2x \] \[\frac{dy}{dx}=2x\sin x+x^2\cos x-2\sin 2x \] Answer: \(\displaystyle 2x\sin x+x^2\cos x-2\sin 2x\)