2.2.2 Answer

2.2.2

1.

(a) $y = \cos^{-1}{x} \Leftrightarrow x = \cos{y}$. Then $1 = -\sin{y}y'$. Now solve for $y'$. Then $y' = -\frac{1}{\sin{y}}$．Pay attention to the range of $y$, express $\sin{y}$ using $x$. Then we have $\sin{y} = \sqrt{1 - \cos^{2}{y}} = \sqrt{1 - x^2}$．Therefore, $y' = \frac{-1}{\sqrt{1 - x^2}}$

(b) $y = \tan^{-1}{x} \Leftrightarrow x = \tan{y}$ implies that $1 = \sec^{2}{y}y'$. Now solve this for $y'$. Then $y' = \frac{1}{\sec^{2}{y}}$．Pay attention to the range of $y$, express $\sec^{2}{y}$ using $x$. Then $\cos^{2}{y} + \sin^{2}{y} = 1$. Thus, $1 + \tan^{2}{y} = \sec^{2}{y}$．Therefore, $y' = \frac{1}{1 + x^2}$

2.

(a) Take logarithm on both sides. Then $${\log{y} = 2\log{x} + \frac{1}{2}\left(\log{(x^3 + 2x + 1) - \log(x^2 -3x + 1) }\right)}$$

Differentiate both sides with respect to $x$.

$${\frac{y'}{y} = \frac{2}{x} + \frac{1}{2}\left(\frac{3x^2 + 2}{x^3 + 2x + 1} - \frac{2x - 3}{x^2 - 3x + 1}\right)}$$

Thus,

$$y' = x^2 \sqrt{\frac{x^3 + 2x + 1}{x^2 - 3x + 1}}\left(\frac{2}{x... ...left(\frac{3x^2 + 2}{x^3 + 2x + 1} - \frac{2x - 3}{x^2 - 3x + 1}\right)\right)$$

(b)Take logarithm on both sides. Then $${\log{y} = \log{x^{x}} = x\log{x}}$$

Now differentiate both sides with respect to $x$.

$${\frac{y'}{y} = \log{x} + x \frac{1}{x} = 1 + \log{x}}$$

Thus,

$$y' = x^{x}(1 + \log{x})$$

(c)Take logarithm on both sides. Then $${\log{y} = \log{(\sin{x})^{x} = x\log{\sin{x}}}}$$

Now differentiate both sides with respect to $x$.

$${\frac{y'}{y} = \log{\sin{x}} + x \frac{\cos{x}}{\sin{x}} }$$

Thus,

$$y' = (\sin{x})^{x}(\log{\sin{x}} + \frac{x\cos{x}}{\sin{x}})$$

(d)Take logarithm on both sides. Then $${\log{y} = \log{x^{1/x}} = \frac{1}{x}\log{x}}$$

Now differentiate both sides with respect to $x$.

$${\frac{y'}{y} = -\frac{1}{x^2}\log{x} + \frac{1}{x}\cdot \frac{1}{x} }$$

Thus,

$$y' = x^{1/x}\left(\frac{1}{x^2} - \frac{\log{x}}{x^2}\right)$$

3.

(a) $${\frac{dy}{dx} = \frac{dy/dt}{dx/dt}}$$ implies that $${\frac{dy}{dx} = \frac{a\cos{t}}{-a\sin{t}} -\frac{\cos{t}}{\sin{t}}}$$

(b) $${\frac{dy}{dx} = \frac{dy/dt}{dx/dt}}$$ implies that $${\frac{dy}{dx} = -\frac{1 - \frac{1}{2}t^{-3/2}}{\frac{1}{2}t^{-1/2} + t^{-2}} = \frac{2t^2 - t^{1/2}}{t^{3/2} + 2}}$$

4.

(a)

$$(x^2(1 + \sqrt{x}))^{\prime} = 2x(1 + \sqrt{x}) + x^2(\frac{1}{2\sqrt{x}}) = 2\,x + \frac{5}{2} x^{3/2}$$

(b)

$$(x^3\tan{2x})^{\prime} = 3x^2 \tan{2x} + x^3(2\sec^{2}{2x}) = 2\,{x^3}\,{{\sec^{2} (2\,x)}} + 3\,{x^2}\,\tan (2\,x)$$

(c)

$$(x \sin^{-1}{x})^{\prime} = \sin^{-1}{x} + x \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{x}{\sqrt{1 - {x^2}}} + \sin^{-1} (x)$$

(d)

$$(\frac{x}{x^2 + 1})^{\prime} = \frac{x^2 + 1 - x(2x)}{(x^2 + 1)^2} = \frac{1 - {x^2}}{\left( 1 + {x^2} \right)^2}$$

(e)

$$(x \sin{x})^{\prime} = \sin (x) + x\,\cos (x)$$

(f)

$$(x\sin^{-1}{x} + \sqrt{1 - x^2})^{\prime} = \sin^{-1}{x} + \frac{x}{\sqrt{1 - x^2}} - \frac{2x}{2\sqrt{1 - x^2}} = \sin^{-1} (x)$$

(g)

$$(\tan^{-1}{(x^2 +1)})^{\prime} = \frac{2\,x}{2 + 2\,{x^2} + {x^4}}$$

(h)

$$(\cos{(\sqrt{2x+1})})^{\prime} = -\frac{\sin ({\sqrt{1 + 2\,x}})}{\sqrt{1 + 2\,x}}$$

(i)

$$\left(\frac{\sin{x} - x\cos{x}}{x\sin{x} + \cos{x}}\right)^{\prime} = \frac{(\cos{x} - \cos{x} + x\sin{x})(x\sin{x} + \cos{x}) - (\sin{x} - x\cos{x})(\sin{x} + x\cos{x} - \sin{x})}{(x\sin{x} + \cos{x})^{2}}$$

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

(j)

$$(e^{2x}\cos{x})^{\prime} = 2e^{2x}\cos{x} - e^{2x} \sin{x} = {e^{2\,x}}\,\left( 2\,\cos (x) - \sin (x) \right)$$

(k)

$$(\log{\vert x + \sqrt{x^2 + A}\vert})^{\prime} = \frac{1 + \frac{... ...+ A}} = \frac{\sqrt{x^2 + A} + x}{\sqrt{x^2 + A}} = \frac{1}{\sqrt{A + {x^2}}}$$