Exercise-5.7, Class 12th, Maths, Chapter 5, NCERT

$\mathrm{<em\; style="font-size:\; var(--wp--preset--font-size--medium);\; letter-spacing:\; -0.1px;\; font-family:\; var(--wp--preset--font-family--manrope);">Find\; the\; second\; order\; derivatives\; of\; the\; functions\; given\; in\; Exercises\; 1\; to\; 10.</em>}$

Q1. $y=x^2+3x+2$

Solution

$$\frac{dy}{dx}=2x+3$$
$$\frac{d^{2}y}{d{x}^2}=2$$

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Q2. $y=x^{20}$

Solution

$$\frac{dy}{dx}=20x^{19}$$
$$\frac{d^{2}y}{d{x}^2}=380x^{18}$$

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Q3. $y=x\cos x$

Solution

Using product rule:

$$\frac{dy}{dx}=\cos x-x\sin x$$

Now differentiate again:

$$\frac{d^{2}y}{d{x}^2}=-\sin x-(\sin x+x\cos x)$$
$$\frac{d^{2}y}{d{x}^2}=-2\sin x-x\cos x$$

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Q4. $y=\log x$

Solution

$$\frac{dy}{dx}=\frac{1}{x}$$
$$\frac{d^{2}y}{d{x}^2}=-\frac{1}{x^2}$$

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Q5. $y=x^3\log x$

Solution

$$\frac{dy}{dx}=3x^2\log x+x^2$$
$$\frac{d^{2}y}{d{x}^2}=6x\log x+5x$$

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Q6. $y=e^x\sin 5x$

Solution

$$\frac{dy}{dx}=e^x(\sin 5x+5\cos 5x)$$
$$\frac{d^{2}y}{d{x}^2}=e^x(-24\sin 5x+10\cos 5x)$$

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Q7. $y=e^{6x}\cos 3x$

Solution$$\frac{dy}{dx}=e^{6x}(6\cos 3x-3\sin 3x)$$
$$\frac{d^{2}y}{d{x}^2}=e^{6x}(27\cos 3x-36\sin 3x)$$

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Q8. $y={\tan }^{-1}x$

Solution$$\frac{dy}{dx}=\frac{1}{1+x^2}$$
$$\frac{d^{2}y}{d{x}^2}=-\frac{2x}{(1+x^2{)}^2}$$

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Q9. $y=\log (\log x)$

Solution$$\frac{dy}{dx}=\frac{1}{x\log x}$$
$$\frac{d^{2}y}{d{x}^2}=-\frac{\log x+1}{x^2(\log x{)}^2}$$

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Q10. $y=\sin (\log x)$

Solution$$\frac{dy}{dx}=\frac{\cos (\log x)}{x}$$
$$\frac{d^{2}y}{d{x}^2}=\frac{-\cos (\log x)-\sin (\log x)}{x^2}$$

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Q11. If $y=5\cos x-3\sin x$, show that $\frac{d^{2}y}{d{x}^2}+y=0$

Solution

$$\frac{dy}{dx}=-5\sin x-3\cos x$$
$$\frac{d^{2}y}{d{x}^2}=-5\cos x+3\sin x$$

Now,$$\frac{d^{2}y}{d{x}^2}+y=(-5\cos x+3\sin x)+(5\cos x-3\sin x)$$

$$\frac{d^{2}y}{d{x}^2}+y=0$$

$$\boxed{{\displaystyle \text{Hence proved}}}$$