SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Limits and Derivatives

73 questions · 73 still being checked

Miscellaneous Exercise 21–30 (part 7 of 7)

  1. Find the derivative of the following functions (it is to be understood that \(\displaystyle a, b, c, d\), \(\displaystyle p, q, r\) and \(\displaystyle s\) are fixed non-zero constants and \(\displaystyle m\) and \(\displaystyle n\) are integers):

    Exercise 21

    sin⁡(x+a)cos⁡x\displaystyle \frac{\sin (x+a)}{\cos x}

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    NCERT’s answer
    $\displaystyle \frac{\cos a}{\cos ^{2} x}$
    Quotient rule, then the cosine subtraction formula.With \(\displaystyle u=\sin(x+a)\) and \(\displaystyle v=\cos x\), and using \(\displaystyle \dfrac{d}{dx}\sin(x+a)=\cos(x+a)\),\[\frac{d}{dx}\frac{\sin(x+a)}{\cos x} =\frac{\cos(x+a)\cos x-\sin(x+a)(-\sin x)}{\cos^{2}x} =\frac{\cos(x+a)\cos x+\sin(x+a)\sin x}{\cos^{2}x}. \]The numerator is \(\displaystyle \cos\big((x+a)-x\big)=\cos a\), a constant.\(\displaystyle \dfrac{\cos a}{\cos^{2}x}=\cos a\,\sec^{2}x\)

  2. Exercise 22

    x4(5sin⁡x−3cos⁡x)\displaystyle x^{4}(5 \sin x-3 \cos x)

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    NCERT’s answer
    $\displaystyle x^{3}(5 x \cos x+3 x \sin x+20 \sin x-12 \cos x)$
    Product rule.Take \(\displaystyle f=x^{4}\) and \(\displaystyle g=5\sin x-3\cos x\), so \(\displaystyle f'=4x^{3}\) and \(\displaystyle g'=5\cos x+3\sin x\).\[\frac{d}{dx}\Big[x^{4}(5\sin x-3\cos x)\Big] =4x^{3}(5\sin x-3\cos x)+x^{4}(5\cos x+3\sin x). \]Take out \(\displaystyle x^{3}\):\(\displaystyle x^{3}\big[(3x+20)\sin x+(5x-12)\cos x\big]\)

  3. Exercise 23

    (x2+1)cos⁡x\displaystyle \left(x^{2}+1\right) \cos x

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    NCERT’s answer
    $\displaystyle -x^{2} \sin x-\sin x+2 x \cos x$
    Product rule.With \(\displaystyle f=x^{2}+1\) and \(\displaystyle g=\cos x\), so \(\displaystyle f'=2x\) and \(\displaystyle g'=-\sin x\),\[\frac{d}{dx}\Big[(x^{2}+1)\cos x\Big]=2x\cos x+(x^{2}+1)(-\sin x). \]\(\displaystyle 2x\cos x-(x^{2}+1)\sin x\)

  4. Exercise 24

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

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    Product rule.Let \(\displaystyle f=ax^{2}+\sin x\) and \(\displaystyle g=p+q\cos x\); then \(\displaystyle f'=2ax+\cos x\) and \(\displaystyle g'=-q\sin x\).\[\frac{d}{dx}\Big[(ax^{2}+\sin x)(p+q\cos x)\Big] =(2ax+\cos x)(p+q\cos x)+(ax^{2}+\sin x)(-q\sin x). \]\(\displaystyle (2ax+\cos x)(p+q\cos x)-q\sin x\,(ax^{2}+\sin x)\)

  5. Exercise 25

    (x+cos⁡x)(x−tan⁡x)\displaystyle (x+\cos x)(x-\tan x)

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    Product rule.Let \(\displaystyle f=x+\cos x\) and \(\displaystyle g=x-\tan x\); then \(\displaystyle f'=1-\sin x\) and \(\displaystyle g'=1-\sec^{2}x\).\[\frac{d}{dx}\Big[(x+\cos x)(x-\tan x)\Big]=(1-\sin x)(x-\tan x)+(x+\cos x)(1-\sec^{2}x). \]Since \(\displaystyle 1+\tan^{2}x=\sec^{2}x\), we have \(\displaystyle 1-\sec^{2}x=-\tan^{2}x\), which tidies the second term.\(\displaystyle (1-\sin x)(x-\tan x)-(x+\cos x)\tan^{2}x\)

  6. Exercise 26

    4x+5sin⁡x3x+7cos⁡x\displaystyle \frac{4 x+5 \sin x}{3 x+7 \cos x}

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    NCERT’s answer
    $\displaystyle \frac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{(3 x+7 \cos x)^{2}}$
    Quotient rule.Let \(\displaystyle u=4x+5\sin x\) and \(\displaystyle v=3x+7\cos x\); then \(\displaystyle u'=4+5\cos x\) and \(\displaystyle v'=3-7\sin x\).\[u'v=(4+5\cos x)(3x+7\cos x)=12x+28\cos x+15x\cos x+35\cos^{2}x, \] \[uv'=(4x+5\sin x)(3-7\sin x)=12x-28x\sin x+15\sin x-35\sin^{2}x. \]Subtracting, the \(\displaystyle 12x\) terms cancel and \(\displaystyle 35\cos^{2}x+35\sin^{2}x=35\):\[u'v-uv'=15x\cos x+28x\sin x+28\cos x-15\sin x+35 . \]\(\displaystyle \dfrac{15x\cos x+28x\sin x+28\cos x-15\sin x+35}{(3x+7\cos x)^{2}}\)

  7. Exercise 27

    x2cos⁡(π4)sin⁡x\displaystyle \frac{x^{2} \cos \left(\dfrac{\pi}{4}\right)}{\sin x}

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    NCERT’s answer
    $\displaystyle \frac{x \cos \dfrac{\pi}{4}(2 \sin x-x \cos x)}{\sin ^{2} x}$
    Pull out the constant, then the quotient rule.\(\displaystyle \cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}\) is a number, not a function of \(\displaystyle x\), so it comes straight out:\[\frac{d}{dx}\left(\frac{x^{2}\cos\dfrac{\pi}{4}}{\sin x}\right)=\frac{1}{\sqrt{2}}\cdot\frac{d}{dx}\left(\frac{x^{2}}{\sin x}\right). \]Now the quotient rule with \(\displaystyle u=x^{2}\), \(\displaystyle v=\sin x\):\[\frac{d}{dx}\frac{x^{2}}{\sin x}=\frac{2x\sin x-x^{2}\cos x}{\sin^{2}x}. \]\(\displaystyle \dfrac{2x\sin x-x^{2}\cos x}{\sqrt{2}\,\sin^{2}x}=\dfrac{x\,(2\sin x-x\cos x)}{\sqrt{2}\,\sin^{2}x}\)

  8. Exercise 28

    x1+tan⁡x\displaystyle \frac{x}{1+\tan x}

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    NCERT’s answer
    $\displaystyle \frac{1+\tan x-x \sec ^{2} x}{(1+\tan x)^{2}}$
    Quotient rule.With \(\displaystyle u=x\) and \(\displaystyle v=1+\tan x\), so \(\displaystyle u'=1\) and \(\displaystyle v'=\sec^{2}x\),\[\frac{d}{dx}\frac{x}{1+\tan x}=\frac{1\cdot(1+\tan x)-x\sec^{2}x}{(1+\tan x)^{2}} . \]\(\displaystyle \dfrac{1+\tan x-x\sec^{2}x}{(1+\tan x)^{2}}\)

  9. Exercise 29

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

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    NCERT’s answer
    $\displaystyle (x+\sec x)\left(1-\sec ^{2} x\right)+(x-\tan x) .(1+\sec x \tan x)$
    Product rule.Let \(\displaystyle f=x+\sec x\) and \(\displaystyle g=x-\tan x\); then \(\displaystyle f'=1+\sec x\tan x\) and \(\displaystyle g'=1-\sec^{2}x\).\[\frac{d}{dx}\Big[(x+\sec x)(x-\tan x)\Big]=(1+\sec x\tan x)(x-\tan x)+(x+\sec x)(1-\sec^{2}x). \]Using \(\displaystyle 1-\sec^{2}x=-\tan^{2}x\), the second term simplifies.\(\displaystyle (1+\sec x\tan x)(x-\tan x)-(x+\sec x)\tan^{2}x\)

  10. Exercise 30

    xsin⁡nx\displaystyle \frac{x}{\sin ^{n} x}

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    NCERT’s answer
    $\displaystyle \frac{\sin x-n x \cos x}{\sin ^{n+1} x}$
    Quotient rule. For \(\displaystyle f(x)=\dfrac{u}{v} \), \(\displaystyle f'(x)=\dfrac{u'v-uv'}{v^{2}} \).Here \(\displaystyle u=x \) and \(\displaystyle v=\sin^{n}x \), so \(\displaystyle u'=1 \).Differentiate \(\displaystyle v \) by the chain rule, treating \(\displaystyle \sin^{n}x \) as \(\displaystyle (\sin x)^{n} \): \[v'=n(\sin x)^{n-1}\cdot\frac{d}{dx}(\sin x)=n\sin^{n-1}x\,\cos x. \]Substituting into the quotient rule, \[\frac{d}{dx}\left(\frac{x}{\sin^{n}x}\right)=\frac{1\cdot\sin^{n}x-x\cdot n\sin^{n-1}x\cos x}{\left(\sin^{n}x\right)^{2}} =\frac{\sin^{n}x-nx\sin^{n-1}x\cos x}{\sin^{2n}x}. \]Take the common factor \(\displaystyle \sin^{n-1}x \) out of the numerator and cancel it against the denominator: \[=\frac{\sin^{n-1}x\left(\sin x-nx\cos x\right)}{\sin^{2n}x} =\frac{\sin x-nx\cos x}{\sin^{n+1}x}. \]Answer: \(\displaystyle \dfrac{\sin x-nx\cos x}{\sin^{n+1}x} \) (valid wherever \(\displaystyle \sin x\neq 0 \)).