Example

Differentiate the following:

1. $y = \arctan (ax^2)$.

   Solution. By XX ($v = ax^2$)

   $$\frac{dy}{dx} = \frac{\frac{d}{dx} (ax^2)}{1 + (ax^2)^2} = \frac{2ax}{1 + a^2x^4}.$$

2. $y = \arcsin(3x - 4x^3)$.

   Solution. By XVIII ( $v = 3x - 4x^3$),

   $$\frac{dy}{dx} = \frac{\frac{d}{dx}(3x - 4x^3)}{\sqrt{1 - (3x - 4x... ...=\frac{3 - 12x^2}{\sqrt{1 - 9x^2 + 24x^4 - 16x^6}} = \frac{3}{\sqrt{1 - x^2}}.$$

3. $y={\rm arcsec}\, \frac{x^2 + 1}{x^2 - 1}$.

   Solution. By XXII ( $v = \frac{x^2 + 1}{x^2 - 1}$),

   $$\frac{dy}{dx} = \frac{\frac{d}{dx} \left ( \frac{x^2 + 1}{x^2 - ... ...)^2}}{\frac{x^2 + 1}{x^2 - 1} \cdot \frac{2x}{x^2 - 1}} = -\frac{2}{x^2 + 1}.$$

4. $\frac{d}{dx} \arcsin \frac{x}{a} = \frac{1}{\sqrt{a^2 - x^2}}$

5. $\frac{d}{dx} {\rm arccot}(x^2 - 5) = \frac{-2x}{1 + (x^2 - 5)^2}$

6. $\frac{d}{dx} \arctan \frac{2x}{1 - x^2} = \frac{2}{1 + x^2}$

7. $\frac{d}{dx} {\rm arccsc}\frac{1}{2x^2 - 1} = \frac{2}{\sqrt{1 - x^2}}$

8. $\frac{d}{dx} \operatorname{arcvers}\, 2x^2 = \frac{2}{\sqrt{1 - x^2}}$

9. $\frac{d}{dx} \arctan \sqrt{1 - x} = -\frac{1}{2 \sqrt{1 - x} (2 - x)}$

10. $\frac{d}{dx} {\rm arccsc}\frac{3}{2x} = \frac{2}{9 - 4x^2}$

11. $\frac{d}{dx} \operatorname{arcvers}\, \frac{2x^2}{1 + x^2} = \frac{2}{1 + x^2}$

12. $\frac{d}{dx} \arctan \frac{x}{a} = \frac{a}{a^2 + x^2}$

13. $\frac{d}{dx} \arcsin \frac{x + 1}{\sqrt{2}} = \frac{1}{\sqrt{1 - 2x - x^2}}$

14. $f(x) = x\sqrt{a^2 - x^2} + a^2 \arcsin \frac{x}{a}$                                        Ans: $f'(x) =\ 2\sqrt{a^2 - x^2}$

15. $f(x) = \sqrt{a^2 - x^2} + a \arcsin \frac{x}{a}$                                        Ans: $f'(x) = \left ( \frac{a - x}{a + x} \right )^{\frac{1}{2}}$

16. $x = r {\rm arcvers}\, \frac{y}{r} - \sqrt{2ry - y^2}$                                                Ans: $\frac{dx}{dy} = \frac{y}{\sqrt{2ry - y^2}}$

17. $\theta = \arcsin (3r - 1)$                                                Ans: $\frac{d\theta}{dr} = \frac{3}{\sqrt{6r - 9r^2}}$

18. $\phi = \arctan \frac{r + a}{1 - ar}$                                                Ans: $\frac{d\phi}{dr} = \frac{1}{1 + r^2}$

19. $s = {\rm arcsec}\frac{1}{\sqrt{1 - t^2}}$                                                Ans: $\frac{ds}{dt} = \frac{1}{\sqrt{1 - t^2}}$

20. $\frac{d}{dx} ( x \arcsin\, x) = \arcsin\, x + \frac{x}{\sqrt{1 - x^2}}$

21. $\frac{d}{d\theta} (\tan \theta \arctan\, \theta) = \sec^2 \theta \arctan\, \theta \frac{\tan \theta}{1 + \theta^2}$

22. $\frac{d}{dt}[\log (\arccos\, t)] = -\frac{1}{\arccos t \sqrt{1 - t^2}}$

23. $f(y) = \arccos (\log\, y)$                                                 Ans: $f'(y) = -\frac{1}{y\sqrt{1 - (\log y)^2}}$

24. $f(\theta) = \arcsin \sqrt{\sin \theta}$                                                Ans: $f'(\theta) = \frac{1}{2} \sqrt{1 + \csc \theta}$

25. $f(\phi) = \arctan \sqrt{\frac{1 - \cos \phi}{1 + \cos \phi}}$                                                Ans: $f'(\phi) = \frac{1}{2}$

26. $p = e^{\arctan\, q}$                                                 Ans: $\frac{dp}{dq} = \frac{e^{\arctan\, q}}{1 + q^2}$

27. $u = \arctan \frac{e^v - e^{-v}}{2}$                                                Ans: $\frac{du}{dv} = \frac{2}{e^v + e^{-v}}$

28. $s = \arccos \frac{e^t - e^{-t}}{e^t + e^{-t}}$                                                Ans: $\frac{ds}{dt} = -\frac{2}{e^v + e^{-v}}$

29. $y = x^{\arcsin x}$                                                Ans: $y' = x^{\arcsin x} \left ( \frac{\arcsin x}{x} + \frac{\log x}{\sqrt{1 - x^2}} \right )$

30. $y = e^{x^x} \arctan\, x$                                                Ans: $y' = e^{x^x} \left [ \frac{1}{1 + x^2} + x^x \arctan x (1 + \log x) \right ]$

31. $y = \arcsin( \sin x )$                                                Ans: $y' = 1$

32. $y = \arctan \frac{4 \sin x}{3 + 5 \cos x}$                                                Ans: $y' = \frac{4}{5 + 3 \cos x}$

33. $y = {\rm arccot}\frac{a}{x} + \log \sqrt{\frac{x - a}{x + a}}$                                                Ans: $y' = \frac{2ax^2}{x^4 - a^4}$

34. $y = \log \left ( \frac{1 + x}{1 - x} \right )^{\frac{1}{4}} - \frac{1}{2} \arctan\, x$                                                Ans: $y' = \frac{x^2}{1 - x^4}$

35. $y = \sqrt{1 - x^2} \arcsin\, x - x$                                                Ans: $y' = - \frac{x \arcsin\, x}{\sqrt{1 - x^2}}$

36. Compute the following derivatives:
    $$\begin{array}{lll} (a)\ \ \frac{d}{dx} \arcsin 2x^2 & (f)\ \... ...a^2} & (o)\ \ \frac{d}{dt} \sqrt{1 - t^2} \arcsin t \end{array}$$

Formulas (5.1) for differentiating a function of a function, and (5.2) for differentiating inverse junctions, have been added to the list of formulas at the beginning of this chapter as XXV and XXVI respectively.

In the next eight examples, first find $\frac{dy}{dv}$ and $\frac{dv}{dx}$ by differentiation and then substitute the results in $\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}$ (by XXV) to find $\frac{dy}{dx}$. (As was pointed out in §5.11, it might be possible to eliminate $v$ between the two given expressions so as to find $y$ directly as a function of $x$, but in most cases the above method is to be preferred.)

In general our results should be expressed explicitly in terms of the independent variable; that is, $\frac{dy}{dx}$ in terms of $x$, $\frac{dx}{dy}$ in terms of $y$, $\frac{d\phi}{d\theta}$ in terms of $\theta$, etc.

1. $y = 2v^2 - 4$, $v = 3x^2 + 1$.

   $\frac{dy}{dv} = 4v$; $\frac{dv}{dx} = 6x$; substituting in XXV, $\frac{dy}{dx} = 4v \cdot 6x = 24x(3x^2 + 1)$.

2. $y = \tan 2v$, $v = \arctan(2x - 1)$.

   $\frac{dy}{dv} = 2 \sec^2 2v$; $\frac{dv}{dx} = \frac{1}{2x^2 - 2x + 1}$; substituting in XXV,

   $$\frac{dy}{dx} = \frac{2 \sec^2 2v}{2x^2 - 2x + 1} = 2 \frac{\tan^2 2v + 1}{2x^2 - 2x + 1} = \frac{2x^2 - 2x + 1}{2(x - x^2)^2}$$
   (since $v = \arctan(2x - 1)$, $\tan v = 2x - 1$, $\tan 2v = \frac{2x - 1}{2x - 2x^2}$).

3. $y = 3 v^2 - 4v + 5, v = 2x^3 - 5$                                                Ans: $\frac{dy}{dx} =\ 72x^5 - 204x^2$

4. $y = \frac{2v}{3v - 2}, v = \frac{x}{2x - 1}$                                                Ans: $\frac{dy}{dx} = \frac{4}{(x - 2)^2}$

5. $y = \log(a^2 - v^2)$                                                Ans: $\frac{dy}{dx} =\ -2 \tan x$

6. $y = \arctan (a + v), v = e^x$                                                Ans: $\frac{dy}{dx} =\ \frac{e^x}{1 + (a + e^x)^2}$

7. $r = e^{2s} + e^s, s = \log(t - t^2)$                                                Ans: $\frac{dr}{dt} =\ 4t^3 - 6t^2 + 1$

In the following examples first find $\frac{dx}{dy}$ by differentiation and then substitute in

$$\frac{dy}{dx} =\ \frac{1}{\frac{dx}{dy}}\ \ \ \ {\rm by\ XXVI}$$

to find $\frac{dy}{dx}$.

1. $x = y\sqrt{1 + y}$                                                Ans: $\frac{dy}{dx} =\ \frac{2\sqrt{1 + y}}{2 + 3y} = \frac{2x}{2y + 3y^2}$

2. $x = \sqrt{1 + \cos\, y}$                                                Ans: $\frac{dy}{dx} =\ -\frac{2 \sqrt{1 + \cos\, y}}{\sin\, y} = -\frac{2}{\sqrt{2 - x^2}}$

3. $x = \frac{y}{1 + \log\, y}$                                                Ans: $\frac{dy}{dx} =\ \frac{(1 + \log y)^2}{\log\, y}$

4. $x = a \log \frac{a + \sqrt{a^2 - y^2}}{y}$                                                Ans: $\frac{dy}{dx} =\ -\frac{y \sqrt{a^2 - y^2}}{a^2}$

5. $x = r {\rm arcvers}\, \frac{y}{r} - \sqrt{2ry - y^2}$                                                Ans: $\frac{dy}{dx} =\ \sqrt{\frac{2r - y}{y}}$

6. Show that the geometrical significance of XXVI is that the tangent makes complementary angles with the two coordinate axes.