SolveItClass 12 · NCERT

NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

137 questions · 137 still being checked

EXERCISE 5.7 1–10 (part 12 of 15)

  1. Find the second order derivatives of the functions given in Exercises $\displaystyle 1$ to $\displaystyle 10$ .

    Exercise 1

    x2+3x+2\displaystyle x^{2}+3 x+2

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle 2$
    Let \(\displaystyle y=x^{2}+3x+2\). The second order derivative is obtained by differentiating \(\displaystyle \frac{dy}{dx}\) once more.By the power rule \(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\), applied term by term, \[\frac{dy}{dx}=2x+3 .\] Differentiating this, \[\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(2x+3)=2 .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=2\ }\] for every real \(\displaystyle x\) (the second derivative of a quadratic is the constant \(\displaystyle 2a\)).

  2. Exercise 2

    x20\displaystyle x^{20}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle 380 x^{18}\)
    Let \(\displaystyle y=x^{20}\). Apply the power rule \(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) twice.First derivative: \[\frac{dy}{dx}=20x^{19} .\] Differentiating \(\displaystyle 20x^{19}\) by the same rule (the constant \(\displaystyle 20\) rides along): \[\frac{d^{2}y}{dx^{2}}=20\cdot 19\,x^{18}=380x^{18} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=380x^{18}\ }\]

  3. Exercise 3

    x⋅cos⁡x\displaystyle x \cdot \cos x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle -x \cos x-2 \sin x\)
    Let \(\displaystyle y=x\cos x\). Use the product rule \(\displaystyle (uv)'=u'v+uv'\).With \(\displaystyle u=x,\ v=\cos x\): \[\frac{dy}{dx}=1\cdot\cos x+x(-\sin x)=\cos x-x\sin x .\] Now differentiate again. The term \(\displaystyle x\sin x\) needs the product rule a second time — this is the step usually dropped: \[\frac{d^{2}y}{dx^{2}}=-\sin x-\big(1\cdot\sin x+x\cos x\big)=-\sin x-\sin x-x\cos x .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-(x\cos x+2\sin x)\ }\]

  4. Exercise 4

    log⁡x\displaystyle \log x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle -\frac{1}{x^{2}}\)
    Let \(\displaystyle y=\log x\), which is defined only for \(\displaystyle x>0\).Standard derivative: \[\frac{dy}{dx}=\frac{1}{x}=x^{-1} .\] Differentiate again by the power rule with \(\displaystyle n=-1\): \[\frac{d^{2}y}{dx^{2}}=(-1)x^{-2}=-\frac{1}{x^{2}} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-\frac{1}{x^{2}},\qquad x>0\ }\]

  5. Exercise 5

    x3log⁡x\displaystyle x^{3} \log x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle x(5+6 \log x)\)
    Let \(\displaystyle y=x^{3}\log x\), defined for \(\displaystyle x>0\). Use the product rule \(\displaystyle (uv)'=u'v+uv'\).With \(\displaystyle u=x^{3},\ v=\log x\): \[\frac{dy}{dx}=3x^{2}\log x+x^{3}\cdot\frac{1}{x}=3x^{2}\log x+x^{2} .\] Differentiate again; the first term needs the product rule once more: \[\frac{d^{2}y}{dx^{2}}=\Big(6x\log x+3x^{2}\cdot\frac{1}{x}\Big)+2x=6x\log x+3x+2x .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=6x\log x+5x=x\,(6\log x+5),\qquad x>0\ }\]

  6. Exercise 6

    exsin⁡5x\displaystyle e^{x} \sin 5 x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle 2 e^{x}(5 \cos 5 x-12 \sin 5 x)\)
    Let \(\displaystyle y=e^{x}\sin 5x\). Use the product rule, and the chain rule on \(\displaystyle \sin 5x\) — the inner factor \(\displaystyle 5\) is what students most often lose.\[\frac{dy}{dx}=e^{x}\sin 5x+e^{x}\cdot 5\cos 5x=e^{x}\big(\sin 5x+5\cos 5x\big).\] Differentiate again by the product rule, differentiating the bracket with the chain rule: \[\frac{d^{2}y}{dx^{2}}=e^{x}\big(\sin 5x+5\cos 5x\big)+e^{x}\big(5\cos 5x-25\sin 5x\big).\] Collecting like terms, \[\frac{d^{2}y}{dx^{2}}=e^{x}\big(10\cos 5x-24\sin 5x\big).\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=e^{x}(10\cos 5x-24\sin 5x)=2e^{x}(5\cos 5x-12\sin 5x)\ }\]

  7. Exercise 7

    e6xcos⁡3x\displaystyle e^{6 x} \cos 3 x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle 9 e^{6 x}(3 \cos 3 x-4 \sin 3 x)\)
    Let \(\displaystyle y=e^{6x}\cos 3x\). Product rule, with the chain rule supplying the inner factors \(\displaystyle 6\) and \(\displaystyle 3\).\[\frac{dy}{dx}=6e^{6x}\cos 3x+e^{6x}(-3\sin 3x)=3e^{6x}\big(2\cos 3x-\sin 3x\big).\] Differentiate again, keeping the factor \(\displaystyle 3\) outside: \[\frac{d^{2}y}{dx^{2}}=3\Big[6e^{6x}\big(2\cos 3x-\sin 3x\big)+e^{6x}\big(-6\sin 3x-3\cos 3x\big)\Big],\] since \(\displaystyle \frac{d}{dx}(2\cos 3x-\sin 3x)=-6\sin 3x-3\cos 3x\). Hence \[\frac{d^{2}y}{dx^{2}}=3e^{6x}\big(12\cos 3x-6\sin 3x-6\sin 3x-3\cos 3x\big)=3e^{6x}\big(9\cos 3x-12\sin 3x\big).\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=9e^{6x}(3\cos 3x-4\sin 3x)\ }\]

  8. Exercise 8

    tan⁡−1x\displaystyle \tan ^{-1} x

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle -\frac{2 x}{\left(1+x^{2}\right)^{2}}\)
    Let \(\displaystyle y=\tan^{-1}x\). Use the standard derivative \(\displaystyle \frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}\).\[\frac{dy}{dx}=\frac{1}{1+x^{2}}=(1+x^{2})^{-1} .\] Differentiate by the chain rule (outer power \(\displaystyle -1\), inner derivative \(\displaystyle 2x\)): \[\frac{d^{2}y}{dx^{2}}=(-1)(1+x^{2})^{-2}\cdot 2x=\frac{-2x}{(1+x^{2})^{2}} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=\frac{-2x}{(1+x^{2})^{2}}\ }\] valid for all real \(\displaystyle x\), since \(\displaystyle 1+x^{2}\) never vanishes.

  9. Exercise 9

    log⁡(log⁡x)\displaystyle \log (\log x)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle -\frac{(1+\log x)}{(x \log x)^{2}}\)
    Let \(\displaystyle y=\log(\log x)\). For this to be defined we need \(\displaystyle \log x>0\), i.e. \(\displaystyle x>1\); the whole answer carries that restriction.Chain rule (outer \(\displaystyle \log\), inner \(\displaystyle \log x\)): \[\frac{dy}{dx}=\frac{1}{\log x}\cdot\frac{1}{x}=\frac{1}{x\log x}=(x\log x)^{-1} .\] Differentiate again by the chain rule, and note that the inner function needs the product rule: \[\frac{d}{dx}(x\log x)=1\cdot\log x+x\cdot\frac{1}{x}=\log x+1 .\] Therefore \[\frac{d^{2}y}{dx^{2}}=-(x\log x)^{-2}\,(1+\log x)=-\frac{1+\log x}{(x\log x)^{2}} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-\frac{(1+\log x)}{(x\log x)^{2}},\qquad x>1\ }\]

  10. Exercise 10

    sin⁡(log⁡x)\displaystyle \sin (\log x)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle -\frac{\sin (\log x)+\cos (\log x)}{x^{2}}\)
    Let \(\displaystyle y=\sin(\log x)\), defined for \(\displaystyle x>0\).Chain rule: \[\frac{dy}{dx}=\cos(\log x)\cdot\frac{1}{x}=\frac{\cos(\log x)}{x} .\] Now the quotient rule \(\displaystyle \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}\) with \(\displaystyle u=\cos(\log x),\ v=x\); the numerator's derivative again needs the chain rule, \(\displaystyle \frac{d}{dx}\cos(\log x)=-\frac{\sin(\log x)}{x}\): \[\frac{d^{2}y}{dx^{2}}=\frac{x\left(-\dfrac{\sin(\log x)}{x}\right)-\cos(\log x)}{x^{2}}=\frac{-\sin(\log x)-\cos(\log x)}{x^{2}} .\]\[\boxed{\ \frac{d^{2}y}{dx^{2}}=-\frac{\sin(\log x)+\cos(\log x)}{x^{2}},\qquad x>0\ }\]