SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Limits and Derivatives

73 questions · 73 still being checked

Miscellaneous Exercise 11–20 (part 6 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 11

    4x−2\displaystyle 4 \sqrt{x}-2

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    NCERT’s answer
    $\displaystyle \frac{2}{\sqrt{x}}$
    Power rule with a fractional index.Write the surd as a power: \(\displaystyle 4\sqrt{x}-2=4x^{1/2}-2 \), valid for \(\displaystyle x>0 \). The rule \(\displaystyle \dfrac{d}{dx}x^{n}=nx^{n-1} \) holds for any rational \(\displaystyle n \), and the constant \(\displaystyle -2 \) differentiates to \(\displaystyle 0 \):\[\frac{d}{dx}\left(4x^{1/2}-2\right)=4\cdot\frac{1}{2}x^{-1/2}-0=2x^{-1/2}=\frac{2}{\sqrt{x}}. \](As a check from first principle, \(\displaystyle \dfrac{\sqrt{x+h}-\sqrt{x}}{h}=\dfrac{1}{\sqrt{x+h}+\sqrt{x}}\to\dfrac{1}{2\sqrt{x}} \), so \(\displaystyle \dfrac{d}{dx}\left(4\sqrt{x}\right)=\dfrac{4}{2\sqrt{x}}=\dfrac{2}{\sqrt{x}} \).)\(\displaystyle \dfrac{2}{\sqrt{x}} \)

  2. Exercise 12

    (ax+b)n\displaystyle (a x+b)^{n}

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    NCERT’s answer
    $\displaystyle n a(a x+b)^{n-1}$
    First principles, using \(\displaystyle \dfrac{d}{dt}t^{n}=nt^{n-1}\).Put \(\displaystyle u=ax+b\). Replacing \(\displaystyle x\) by \(\displaystyle x+h\) turns \(\displaystyle u\) into \(\displaystyle u+ah\), so\[\frac{d}{dx}(ax+b)^{n}=\lim_{h\to 0}\frac{\big(u+ah\big)^{n}-u^{n}}{h}. \]Substitute \(\displaystyle k=ah\). Since \(\displaystyle a\neq 0\), \(\displaystyle k\to 0\) exactly when \(\displaystyle h\to 0\), and \(\displaystyle h=\dfrac{k}{a}\):\[=a\lim_{k\to 0}\frac{(u+k)^{n}-u^{n}}{k}=a\cdot nu^{\,n-1}. \]The limit in the middle is precisely the derivative of \(\displaystyle t^{n}\) at \(\displaystyle t=u\).\(\displaystyle \dfrac{d}{dx}(ax+b)^{n}=na(ax+b)^{n-1}\)

  3. Exercise 13

    (ax+b)n(cx+d)m\displaystyle (a x+b)^{n}(c x+d)^{m}

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    NCERT’s answer
    $\displaystyle (a x+b)^{n-1}(c x+d)^{m-1}[m c(a x+b)+n a(c x+d)]$
    Product rule, using the result of the previous question.From \(\displaystyle \dfrac{d}{dx}(ax+b)^{n}=na(ax+b)^{n-1}\) we also get \(\displaystyle \dfrac{d}{dx}(cx+d)^{m}=mc(cx+d)^{m-1}\).Apply \(\displaystyle (fg)'=f'g+fg'\) with \(\displaystyle f=(ax+b)^{n}\), \(\displaystyle g=(cx+d)^{m}\):\[\frac{d}{dx}\Big[(ax+b)^{n}(cx+d)^{m}\Big] =na(ax+b)^{n-1}(cx+d)^{m}+mc\,(ax+b)^{n}(cx+d)^{m-1}. \]Take out the common factor \(\displaystyle (ax+b)^{n-1}(cx+d)^{m-1}\):\(\displaystyle (ax+b)^{n-1}(cx+d)^{m-1}\Big[na(cx+d)+mc(ax+b)\Big]\)

  4. Exercise 14

    sin⁡(x+a)\displaystyle \sin (x+a)

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    NCERT’s answer
    $\displaystyle \cos (x+a)$
    Expand with the addition formula, then differentiate term by term.Here \(\displaystyle a\) is a constant, so \(\displaystyle \sin a\) and \(\displaystyle \cos a\) are constants:\[\sin(x+a)=\sin x\cos a+\cos x\sin a . \]Differentiating, with \(\displaystyle \cos a\) and \(\displaystyle \sin a\) carried along as constant multipliers,\[\frac{d}{dx}\sin(x+a)=\cos a\cdot\cos x+\sin a\cdot(-\sin x)=\cos x\cos a-\sin x\sin a . \]The right-hand side is the expansion of \(\displaystyle \cos(x+a)\).\(\displaystyle \dfrac{d}{dx}\sin(x+a)=\cos(x+a)\)

  5. Exercise 15

    cosec⁡xcot⁡x\displaystyle \operatorname{cosec} x \cot x

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    NCERT’s answer
    $\displaystyle -\operatorname{cosec}^{3} x-\operatorname{cosec} x \cot ^{2} x$
    Product rule.Use \(\displaystyle \dfrac{d}{dx}\operatorname{cosec}x=-\operatorname{cosec}x\cot x\) and \(\displaystyle \dfrac{d}{dx}\cot x=-\operatorname{cosec}^{2}x\).With \(\displaystyle f=\operatorname{cosec}x\) and \(\displaystyle g=\cot x\), \(\displaystyle (fg)'=f'g+fg'\) gives\[\frac{d}{dx}\big(\operatorname{cosec}x\cot x\big) =\big(-\operatorname{cosec}x\cot x\big)\cot x+\operatorname{cosec}x\big(-\operatorname{cosec}^{2}x\big) \] \[=-\operatorname{cosec}x\cot^{2}x-\operatorname{cosec}^{3}x . \]\(\displaystyle -\operatorname{cosec}x\left(\cot^{2}x+\operatorname{cosec}^{2}x\right)\)

  6. Exercise 16

    cos⁡x1+sin⁡x\displaystyle \frac{\cos x}{1+\sin x}

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    NCERT’s answer
    $\displaystyle \frac{-1}{1+\sin x}$
    Quotient rule.\(\displaystyle \left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}\), with \(\displaystyle u=\cos x\) and \(\displaystyle v=1+\sin x\):\[\frac{d}{dx}\frac{\cos x}{1+\sin x} =\frac{(-\sin x)(1+\sin x)-\cos x\,(\cos x)}{(1+\sin x)^{2}} =\frac{-\sin x-\sin^{2}x-\cos^{2}x}{(1+\sin x)^{2}} . \]Since \(\displaystyle \sin^{2}x+\cos^{2}x=1\), the numerator is \(\displaystyle -\sin x-1=-(1+\sin x)\), and one factor of \(\displaystyle 1+\sin x\) cancels.\(\displaystyle -\dfrac{1}{1+\sin x}\)

  7. Exercise 17

    sin⁡x+cos⁡xsin⁡x−cos⁡x\displaystyle \frac{\sin x+\cos x}{\sin x-\cos x}

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    NCERT’s answer
    $\displaystyle \frac{-2}{(\sin x-\cos x)^{2}}$
    Quotient rule.Take \(\displaystyle u=\sin x+\cos x\) and \(\displaystyle v=\sin x-\cos x\), so \(\displaystyle u'=\cos x-\sin x=-v\) and \(\displaystyle v'=\cos x+\sin x=u\). Then\[\frac{d}{dx}\frac{u}{v}=\frac{u'v-uv'}{v^{2}}=\frac{-v\cdot v-u\cdot u}{v^{2}}=\frac{-\big[(\sin x-\cos x)^{2}+(\sin x+\cos x)^{2}\big]}{(\sin x-\cos x)^{2}} . \]The bracket expands to \(\displaystyle (1-2\sin x\cos x)+(1+2\sin x\cos x)=2\).\(\displaystyle -\dfrac{2}{(\sin x-\cos x)^{2}}\)

  8. Exercise 18

    sec⁡x−1sec⁡x+1\displaystyle \frac{\sec x-1}{\sec x+1}

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    NCERT’s answer
    $\displaystyle \frac{2 \sec x \tan x}{(\sec x+1)^{2}}$
    Quotient rule.With \(\displaystyle u=\sec x-1\) and \(\displaystyle v=\sec x+1\), both have the same derivative \(\displaystyle u'=v'=\sec x\tan x\). So\[\frac{d}{dx}\frac{\sec x-1}{\sec x+1} =\frac{\sec x\tan x(\sec x+1)-(\sec x-1)\sec x\tan x}{(\sec x+1)^{2}} \] \[=\frac{\sec x\tan x\big[(\sec x+1)-(\sec x-1)\big]}{(\sec x+1)^{2}} =\frac{2\sec x\tan x}{(\sec x+1)^{2}} . \](Multiplying numerator and denominator of the original by \(\displaystyle \cos x\) first gives \(\displaystyle \dfrac{1-\cos x}{1+\cos x}\), and differentiating that gives \(\displaystyle \dfrac{2\sin x}{(1+\cos x)^{2}}\) — the same answer in another dress.)\(\displaystyle \dfrac{2\sec x\tan x}{(\sec x+1)^{2}}\;=\;\dfrac{2\sin x}{(1+\cos x)^{2}}\)

  9. Exercise 19

    sin⁡nx\displaystyle \sin ^{n} x

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    NCERT’s answer
    $\displaystyle n \sin ^{\mathrm{n}-1} x \cos x$
    Induction on \(\displaystyle n\), using the product rule.Claim: \(\displaystyle \dfrac{d}{dx}\sin^{n}x=n\sin^{n-1}x\cos x\) for every positive integer \(\displaystyle n\).Base case \(\displaystyle n=1\): \(\displaystyle \dfrac{d}{dx}\sin x=\cos x=1\cdot\sin^{0}x\cos x\). True.Inductive step: suppose the claim holds for \(\displaystyle n\). Writing \(\displaystyle \sin^{n+1}x=\sin^{n}x\cdot\sin x\), the product rule gives\[\frac{d}{dx}\sin^{n+1}x=\big(n\sin^{n-1}x\cos x\big)\sin x+\sin^{n}x\cos x =n\sin^{n}x\cos x+\sin^{n}x\cos x=(n+1)\sin^{n}x\cos x, \]which is the claim for \(\displaystyle n+1\). By induction it holds for all \(\displaystyle n\).\(\displaystyle \dfrac{d}{dx}\sin^{n}x=n\sin^{n-1}x\cos x\)

  10. Exercise 20

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

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    NCERT’s answer
    $\displaystyle \frac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^{2}}$
    Quotient rule.Let \(\displaystyle u=a+b\sin x\) and \(\displaystyle v=c+d\cos x\); then \(\displaystyle u'=b\cos x\) and \(\displaystyle v'=-d\sin x\).\[\frac{d}{dx}\frac{a+b\sin x}{c+d\cos x} =\frac{b\cos x\,(c+d\cos x)-(a+b\sin x)(-d\sin x)}{(c+d\cos x)^{2}} . \]Expand the numerator:\[bc\cos x+bd\cos^{2}x+ad\sin x+bd\sin^{2}x =ad\sin x+bc\cos x+bd, \]using \(\displaystyle \sin^{2}x+\cos^{2}x=1\).\(\displaystyle \dfrac{ad\sin x+bc\cos x+bd}{(c+d\cos x)^{2}}\)