# Elements of the Differential and Integral Calculus/Chapter V part 2

44. Differentiation of a logarithm.

Let $y = \log_a v$.[1]

Differentiating by the General Rule, p. 29 [§ 31], considering $v$ as the independent variable, we have

 FIRST STEP. $\ y + \Delta y$ $= \log_a(v + \Delta v)\$. SECOND STEP. $\ \Delta y$ $= \log_a(v + \Delta v) - \log_a v\$[2] $= \log_a \left ( \frac{v + \Delta v}{v} \right ) = \log_a \left ( 1 + \frac{\Delta v}{v} \right )$. [By 8, p. 1 [§ 1]] THIRD STEP. $\frac{\Delta y}{\Delta x}$ $= \frac{1}{\Delta v} \log_a \left ( 1 + \frac{\Delta v}{v} \right ) = \log_a \left ( 1 + \frac{\Delta v}{v} \right )^{\frac{1}{\Delta v}}$ $= \frac{1}{v} \log_a \left ( 1 + \frac{\Delta v}{v} \right )^{\frac{v}{\Delta v}}$. [Dividing the logarithm by $v$ and at the same time multiplying the exponent of the parenthesis by $v$ changes the form of the expression but not its value (see 9, p. 1 [§ 1]).] FOURTH STEP. $\frac{dy}{dv}$ $= \frac{1}{v} \log_a e$. [When $\Delta v \dot= 0, \frac{\Delta v}{v} \dot= 0$. Therefore $\lim{\Delta v \to 0} \left ( 1 + \frac{\Delta v}{v} \right )^{\frac{v}{\Delta v}} = e$, from p. 22 (§ 23), placing $x = \frac{\Delta v}{v}$.] Hence (A) $\frac{dy}{dv}$ $= \frac{d}{dv} \left ( \log_a v \right ) = \log_a e \cdot \frac{1}{v}$.

Since $v$ is a function of $x$ and it is required to differentiate $\log_a v$ with respect to $x$, we must use formula (A), § 42, for differentiating a function of a function, namely,

 $\frac{dy}{dx}$ $= \frac{dy}{dv} \cdot \frac{dv}{dx}$. Substituting value of $\frac{dy}{dv}$ from (A), we get $\frac{dy}{dx}$ $= \log_a e \cdot \frac{1}{v} \cdot \frac{dv}{dx}$. VIII ∴ $\frac{d}{dx} (\log_a x)$ $= \log_s e \cdot \frac{\frac{dv}{dx}}{v}$. When $a = e,\ \log_a e = log_e e = 1$, and VIII becomes VIIIa $\frac{d}{dx} (\log v)$ $= \frac{\frac{dv}{dx}}{v}$.

The derivative of the logarithm of a function is equal to the product of the modulus[3] of the system of logarithms and the derivative of the function, divided by the function.

45. Differentiation of the simple exponential function.

 Let $\ y$ $= a^v. \qquad a > 0\$ Taking the logarithm of both sides to the base $e$, we get $\ \log y$ $= v \log a\$, or $\ v$ $= \frac{\log y}{\log a}$ $= \frac{1}{\log a} \cdot \log y$. Differentiate with respect to $y$ by formula VIIIa, $\frac{dv}{dy}$ $= \frac{1}{\log a} \cdot \frac{1}{y}$; and from (C), § 43, relating to inverse functions, we get $\frac{dy}{dv}$ $= \log a \cdot y$, or, (A) $\frac{dy}{dv}$ $= \log a \cdot a^v$, Since $v$ is a function of $x$ and it is required to differentiate $a^v$ with respect to $x$, we must use formula (A), § 42, for differentiating a function of a function, namely, $\frac{dy}{dx}$ $= \frac{dy}{dv} \cdot \frac{dv}{dx}$. Substituting the value of $\frac{dy}{dx}$ from (A), we get $\frac{dy}{dx}$ $= \log a \cdot a^v \cdot \frac{dv}{dx}$. IX ∴ $\frac{d}{dx} (a^v)$ $= \log a \cdot a^v \cdot \frac{dv}{dx}$. When $a = e,\ \log a = \log e = 1$, and IX becomes IXa $\frac{d}{dx} (e^v)$ $= e^v \frac{dv}{dx}$.

The derivative of a constant with a variable exponent is equal to the product of the natural logarithm of the constant, the constant with the variable exponent, and the derivative of the exponent.

46. Differentiation of the general exponential function.

 Let $\ y$ $= u^v\$.[4] Taking the logarithm of both sides to the base $e$, $\ \log_e y$ $= v log_e u\$, or, $\ y$ $= e^{v \log u}\$. Differentiating by formula Ixa, $\frac{dy}{dx}$ $= e^{v \log u} \frac{d}{dx} (v \log u)$ $= e^{v \log u} \left ( \frac{v}{u} \frac{du}{dx} + \log u \frac{dv}{dx} \right )$ by V $= u^v \left ( \frac{v}{u} \frac{du}{dx} + \log u \frac{dv}{dx} \right )$ X ∴ $\frac{d}{dx}(u^v)$ $= vu^{v - 1}\frac{du}{dx} + \log u \cdot u^v \frac{dv}{dx}$.

'The derivative of a function with a variable exponent is equal to the sum of the two results obtained by first differentiating by VI, regarding the exponent as constant,. and again differentiating by IX, regarding the function as constant.

Let v = n, any constant; then X reduces to

$\frac{d}{dx}(u^n) = nu^{n - 1} \frac{du}{dx}$.

But this is the form differentiated in § 40; therefore VI holds true for any value of $n$.

 ILLUSTRATIVE EXAMPLE 1. Differentiate $y = log(x^2 + a)$. Solution. $\frac{dy}{dx}$ $= \frac{\frac{d}{dx}(x^2 + a)}{x^2 + a}$ by VIIIa [v = x^2 + a] $= \frac{2x}{x^2 + a}$ Ans. ILLUSTRATIVE EXAMPLE 2. Differentiate $y = \log \sqrt{1 - x^2}$. Solution. $\frac{dy}{dx}$ $= \frac{\frac{d}{dx}(1 - x^2)^{\frac{1}{2}}}{(1 - x^2)^{\frac{1}{2}}}$ by VIIIa $= \frac{\frac{1}{2} (1 - x^2)^{-\frac{1}{2}}(-2x)}{(1 - x^2)^{\frac{1}{2}}}$ by VI $= \frac{x}{x^2 - 1}$. Ans. ILLUSTRATIVE EXAMPLE 3. Differentiate $y = a^{3x^2}$. Solution: $\frac{dy}{dx}$ $= \log a \cdot a^{3x^2} \frac{d}{dx}(3 x^2)$ by IX $= 6x \log a \cdot a^{3x^2}$ Ans. ILLUSTRATIVE EXAMPLE 4. Differentiate $y = be^{c^2 + x^2}$. Solution. $\frac{dy}{dx}$ $b\frac{d}{dx} \left ( e^{c^2 + x^2} \right )$. by IV $= be^{c^2 + x^2} \frac{d}{dx} (c^2 + x^2)$ by IXa $= 2bxe^{c^2 + x^2}$. Ans. ILLUSTRATIVE EXAMPLE 5. Differentiate $y = xe^x$. Solution: $\frac{dy}{dx}$ $= e^x x^{e^x - 1} \frac{d}{dx} (x) + x^{e^x} \log x \frac{d}{dx} (e^x)$ by X $= e^x x^{e^x - 1} + x^{e^x} \log x \cdot e^x$ $= e^x x^{e^x} \left ( \frac{1}{x} + \log x \right )$ Ans.

47. Logarithmic differentiation. Instead of applying VIII and VIIIa at once in differentiating logarithmic functions, we may sometimes simplify the work by first making use of one of the formulas 7-10 on p. 1 [§ 1]. Thus above Illustrative Example 2 may be solved as follows:

 ILLUSTRATIVE EXAMPLE 1. Differentiate $y = \log \sqrt{1 - x^2}$. Solution. By using 10, p. 1, we may write this in a form free from radicals as follows: $y$ $= \frac{1}{2} \log (1 - x^2)$. Then $\frac{dy}{dx}$ $= \frac{1}{2} \frac{\frac{d}{dx} (1 - x^2)}{1 - x^2}$ by VIIIa $= \frac{1}{2} \cdot \frac{-2}{1 - x^2} = \frac{x}{x^2 - 1}$. Ans. ILLUSTRATIVE EXAMPLE 2. Differentiate $y = log \sqrt{\frac{1 + x^2}{1 - x^2}}$. Solution. Simplifying by means of 10 and 8, p. 1 [§ 1], $y$ $= \frac{1}{2} [ \log (1 + x^2) - \log (1 - x^2) ]$. $\frac{dy}{dx}$ $= \frac{1}{2} \left [ \frac{\frac{d}{dx} (1 + x^2)}{1 + x^2} - \frac{\frac{d}{dx} (1 - x^2)}{1 - x^2} \right ]$ by VIIIa, etc. $= \frac{x}{1 + x^2} + \frac{x}{1 - x^2} = \frac{2x}{1 - x^4}$. Ans.

In differentiating an exponential function, especially a variable with a variable exponent, the best plan is first to take the logarithm of the function and then differentiate. Thus Illustrative Example 5, p. 50 [§ 46], is solved more elegantly as follows:

 ILLUSTRATIVE EXAMPLE 3. Differentiate $y = x^{e^x}$. Solution. Taking the logarithm of both sides, $\log y = e^x \log x$. By 9, p. 1 [§ 1] Now differentiate both sides with respect to $x$. $\frac{\frac{dy}{dx}}{y}$ $= e^x \frac{d}{dx} (\log x) + \log x \frac{d}{dx} (e^x)$ by VIII and V $= e^x \cdot \frac{1}{x} + \log x \cdot e^x$, or, $\frac{dy}{dx}$ $= e^x \cdot y \left ( \frac{1}{x} \log x \right )$ $= e^x x^{e^x} \left ( \frac{1}{x} + \log x \right )$. Ans.

ILLUSTRATIVE EXAMPLE 4. Differentiate $y = (4x^2 - 7)^{2 + \sqrt{x^2 - 5}}$.

Solution. Taking the logarithm of both sides,

$\log y = (2 + \sqrt{x^2 - 5}) \log (4x^2 - 7)$.

Differentiating both sides with respect to $x$,

 $\frac{1}{y} \frac{dy}{dx}$ $= (2 + \sqrt{x^2 - 5}) \frac{8x}{4x^2 - 7} + \log(4x^2 - 7) \cdot \frac{x}{\sqrt{x^2 - 5}}$. $\frac{dy}{dx}$ $= x(4x^2 - 7)^{2 + \sqrt{x^2 - 5}} \left [ \frac{8(2 + \sqrt{x^2 - 5})}{4x^2 - 7} + \frac{\log (4x^2 - 7)}{\sqrt{x^2 - 5}} \right ]$. Ans.

In the case of a function consisting of a number of factors it is sometimes convenient to take the logarithm before differentiating. Thus,

ILLUSTRATIVE EXAMPLE 5. Differentiate $y = \sqrt{\frac{(x - 1)(x - 2)}{(x - 3)(x - 4)}}$.

Solution. Taking the logarithm of both sides,

$\log y = \frac{1}{2} [\log (x -1) + \log (x - 2) - \log(x - 3) - \log(x - 4)]$.

Differentiating both sides with respect to $x$,

 $\frac{1}{y} \frac{dy}{dx}$ $= \frac{1}{2} \left [ \frac{1}{x - 1} + \frac{1}{x - 2} - \frac{1}{x - 3} - \frac{1}{x - 4} \right ]$ $= -\frac{2x^2 - 10x + 11}{(x - 1)(x - 2)(x - 3)(x - 4)}$, or, $\frac{dy}{dx}$ $= -\frac{2x^2 - 10x - 11}{(x - 1)^{\frac{1}{2}} (x - 2)^{\frac{1}{2}} ( x - 3)^{\frac{3}{2}} (x - 4)^{\frac{3}{2}}}$. Ans.

EXAMPLES

Differentiate the following:

 1. $y = \log (x + a)$. $\frac{dy}{dx} = \frac{1}{x + a}$. 2. $y = \log (ax + b)$. $\frac{dy}{dx} = \frac{a}{ax + b}$. 3. $y = \log \frac{1 + x^2}{1 - x^2}$. $\frac{dy}{dx} = \frac{4x}{1 - x^4}$. 4. $y = \log (x^2 + x)$ $y' = \frac{2x + 1}{x^2 + x}$. 5. $y = \log (x^3 - 2x + 5)$. $y' = \frac{3x^2 - 2}{x^3 - 2x + 5}$. 6. $y = \log_a (2x + x^3)$. $y' = \log_a e \cdot \frac{2 + 3x^2}{2x + x^3}$. 7. $y = x \log x$. $y' = \log x + 1$. 8. $f(x) = \log x^3$. $f'(x) = \frac{3}{x}$. 9. $f(x) = \log^3 x$. $f'(x) = \frac{3 \log^2 x}{x}$. HINT. $\log^3 x = (\log x)^3$. Use first VI, $v = \log x, n = 3$; and then VIIIa. 10. $f(x) = \log \frac{a + x}{a - x}$. $f'(x) = \frac{2a}{a^2 - x^2}$. 11. $f(x) = \log (x + \sqrt{1 + x^2})$. $f'(x) = \frac{1}{\sqrt{1 + x^2}}$.

12. $\frac{d}{dx} e^{ax} = ae^{ax}$.

13. $\frac{d}{dx} e^{4x + 5} = 4e^{4x + 5}$.

14. $\frac{d}{dx} a^{3x} = 3a^{3x} \log a$.

15. $\frac{d}{dt} \log(3 - 2t^2) = \frac{4t}{2t^2 - 3}$.

16. $\frac{d}{dy} \log \frac{1 + y}{1 - y} = \frac{2}{1 - y^2}$.

17. $\frac{d}{dx}e^{b^2 + x^2} = 2xe^{b^2 + x^2}$.

18. $\frac{d}{d\theta} a^{\log a} = \frac{1}{\theta} a^{\log \theta} \log a$.

19. $\frac{d}{ds}b^{s^2} = 2x \log b \cdot b^{s^2}$.

20. $\frac{d}{dv} ae^{\sqrt{v}} = \frac{ae^{\sqrt{v}}}{2\sqrt{v}}$.

21. $\frac{d}{dx} a^{e^x} = \log a \cdot a^{e^x} \cdot e^x$.

 22. $y = 7^{x^2 + 2x}$. $y' = 2\log 7 \cdot (x + 1) 7^{x^2 + 2x}$. 23. $y = c^{a^2 - x^2}$. $y' = -2x \log c \cdot c^{a^2 - x^2}$. 24. $y = \log \frac{e^x}{1 + e^x}$. $\frac{dy}{dx} = \frac{1}{1 + e^x}$.

25. $\frac{d}{dx} \left [ e^x ( 1- x^2 \right ] = e^x (1 - 2x - x^2)$.

26. $\frac{d}{dx} \left ( \frac{e^x - 1}{e^x + 1} \right ) = \frac{2e^x}{(e^x + 1)^2}$

27. $\frac{d}{dx} \left ( x^2 e^{ax} \right ) = xe^{ax}(ax + 2)$.

 28. $y = \frac{a}{2} (e^{\frac{x}{a}} - e^{-\frac{x}{a}}).$ $\frac{dy}{dx} = \frac{1}{2} (e^{\frac{x}{a}} + e^{-\frac{x}{a}})$. 29. $y = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. $\frac{dy}{dx} = \frac{4}{(e^x + e^{-x}))^2}$. 30. $y = x^n a^x$. $y' = a^x x^{n - 1}(n + x \log a)$. 31. $y = x^x$. $y' = x^x(\log x + 1)$. 32. $y = x^{\frac{1}{x}}$. $y' = \frac{x^{\frac{1}{x}} (1 - \log x)}{x^2}$. 33. $y = x^{\log x}$. $y' = \log x^2 \cdot x^{\log x - 1}$. 34. $f(y) = \log y \cdot e^y$. $f'(y) = e^y \left ( \log y + \frac{1}{y} \right )$. 35. $f(s) = \frac{\log s}{e^s}$. $f'(s) = \frac{1 - s \log s}{s e^s}$ 36. $f(x) = \log (\log x)$. $f'(x) = \frac{1}{x \log x}$. 37. $F(x) = \log^4 (\log x)$ $F'(x) = \frac{4 \log^3 (\log x)}{x \log x}$. 38. $\phi(x) = \log(\log^4 x)$. $\phi'(x) = \frac{4}{x \log x}$. 39. $\psi(y) = \log \sqrt{\frac{1 + y}{1 - y}}$. $\psi'(y) = \frac{1}{1 - y^2}$. 40. $f(x) = \log \frac{\sqrt{x^2 + 1} - x}{\sqrt{x^1 + 1} + x}$. $f'(x) = -\frac{2}{\sqrt{1 + x^2}}$. HINT. First rationalize the denominator. 41. $y = x^{\frac{1}{\log x}}$. $\frac{dy}{dx} = 0$. 42. $y = e^{x^x}$. $\frac{dy}{dx} = e^{x^x}(1 + \log x)x^x$. 43. $y = \frac{c^x}{x^x}$ $\frac{dy}{dx} = \left ( \frac{c}{x} \right )^x \left ( \log \frac{c}{x} - 1 \right )$. 44. $y = \left ( \frac{x}{n} \right )^{nx}$. $\frac{dy}{dx} = n \left ( \frac{x}{n} \right )^{nx} \left ( 1 + \log \frac{x}{n} \right )$. 45. $w = v^{e^v}$. $\frac{dw}{dv} = v^{e^v} e^v \left ( \frac{1 + v \log v}{v} \right )$. 46. $z = \left ( \frac{a}{t} \right )^t$. $\frac{dz}{dt} = \left ( \frac{a}{t} \right )^t (\log a - \log t - 1)$. 47. $y = x^{x^n}$. $\frac{dy}{dx} = x^{x^n + n - 1}(n \log x + 1)$. 48. $y = x^{x^x}$. $\frac{dy}{dx} = x^{x^x} x^x \left ( \log x + \log^2 x + \frac{1}{x} \right )$. 49. $y = a^{\frac{1}{\sqrt{a^2 - x^2}}}$. $\frac{dy}{dx} = \frac{xy \log a}{(a^2 - x^2)^{\frac{3}{2}}}$.

50. Differentiate the following functions:

 (a) $\frac{d}{dx} x^2 \log x$. (f) $\frac{d}{dx} e^x \log x$. (k) $\frac{d}{dx} \log (a^x + b^x)$. (b) $\frac{d}{dx} (e^{2x} - 1)^4$. (g) $\frac{d}{dx} x^3 3^x$ (l) $\frac{d}{dx} \log_10 (x^2 + 5x)$. (c) $\frac{d}{dx} \log \frac{3x + 1}{x + 3}$. (h) $\frac{d}{dx} \frac{1}{x \log x}$. (m) $\frac{d}{dx} \frac{2 + x^2}{e^{3x}}$. (d) $\frac{d}{dx} \log \frac{1 - x^2}{\sqrt{1 + x}}$. (i) $\frac{d}{dx} \log x^3 \sqrt{1 + x^2}$. (n) $\frac{d}{dx} (x^2 + a^2) e^{x^2 + a^2}$. (e) $\frac{d}{dx} x^{\sqrt{x}}$. (j) $\frac{d}{dx} \left ( \frac{1}{x} \right )^x$. (o) $\frac{d}{dx} (x^2 + 4)^x$.
 51. $y = \frac{(x + 1)^2}{(x + 2)^3 (x + 3)^4}$. $\frac{dy}{dx} = -\frac{(x + 1)(5x^2 + 14x + 5)}{(x + 2)^4 (x + 3)^5}$. HINT. Take logarithm of both sides before differentiating in this and the following examples. 52. $y = \frac{((x - 1)^{\frac{5}{2}}}{(x - 2)^{\frac{3}{4}}(x - 3)^{\frac{7}{3}}}$. $\frac{dy}{dx} = -\frac{(x - 1)^{\frac{3}{2}}(7x^2 + 30x - 97)}{12(x - 2)^{\frac{7}{4}}(x - 3)^{\frac{10}{3}}}$. 53. $\frac{dy}{dx} = x \sqrt{1 - x} (1 + x)$. $\frac{dy}{dx} = \frac{2 + x - 5x^2}{2\sqrt{1 - x}}$. 54. $y = \frac{x(1 + x^2)}{\sqrt{1 - x^2}}$ $\frac{dy}{dx} = \frac{1 + 3x^2 - 2x^4}{(1 - x^2}^{\frac{3}{2}}$. 55. $y = x^5(a + 3x)^3(a - 2x)^2$. $\frac{dy}{dx} = 5x^4(a + 3x)^2(a - 2x)(a^2 + 2ax - 12x^2)$.

48. Differentiation of $\sin v$.

 Let $y$ $= \sin v$ By General Rule, p. 29 [§ 31], considering $v$ as the independent variable, we have FIRST STEP. $y + \Delta y$ $= \sin(v + \Delta v)$. SECOND STEP. $\Delta y$ $= \sin(v + \Delta v) - \sin v$[5] $= 2 \cos \left ( v + \frac{\Delta v}{2} \right ) \cdot \sin \frac{\Delta v}{2}$.[6] THIRD STEP. $\frac{\Delta y}{\Delta v}$ $= \cos \left ( v + \frac{\Delta v}{2} \right ) \left ( \frac{\sin \frac{\Delta v}{2}}{\frac{\Delta v}{2}} \right )$. FOURTH STEP. $\frac{dy}{dx}$ $= \cos v$.

[ Since $\lim_{\Delta v \to 0} \left ( \frac{\sin \frac{\Delta v}{2}}{\frac{\Delta v}{2}} \right ) = 1,$ by § 22, p. 21, and $\lim_{\Delta v \to 0} \cos \left ( v + \frac{\Delta v}{2} \right ) = \cos v$ ].

Since $v$ is a function of $x$ and it is required to differentiate $\sin v$ with respect to $x$, we must use formula (A), § 42, for differentiating a function of a function, namely,

 $\frac{dy}{dx}$ $= \frac{dy}{dv} \cdot \frac{dv}{dx}$. Substituting value $\frac{dy}{dx}$ from Fourth Step, we get $\frac{dy}{dx}$ $= \cos v \frac{dv}{dx}$. XI ∴ $\frac{d}{dx} (\sin v)$ $= \cos v \frac{dv}{dx}$.

The statement of the corresponding rules will now be left to the student.

49. Differentiation of $\cos v$.

 Let $y$ $= \cos v$. By 29, p. 2 [§ 1], this may be written $y$ $= \sin \left ( \frac{\pi}{2} - v \right )$. Differentiating by formula XI, $\frac{dy}{dx}$ $= \cos \left ( \frac{\pi}{2} - v \right ) \frac{d}{dx} \left ( \frac{\pi}{2} - v \right )$ $= \cos \left ( \frac{\pi}{2} - v \right ) \left ( -\frac{d}{dx} \right )$ $= -\sin x \frac{dv}{dx}$. [ Since $\cos \left ( \frac{\pi}{2} \right ) = \sin v$, by 29, p. 2.] XII ∴ $\frac{d}{dx} (\cos v)$ $= -\sin v \frac{dv}{dx}$.

50. Differentiation of $\tan v$.

 Let $y$ $= \tan v$. By 27, p. 2 [§ 1], this may be written $\frac{dy}{dx}$ $= \frac{\cos v \frac{d}{dx}(\sin v) - \sin v \frac{d}{dx}(\cos v)}{\cos^2 v}$ $= \frac{\cos^2 v \frac{dv}{dx} + \sin^2 v \frac{dv}{dx}}{\cos^2 v}$ $= \frac{\frac{dv}{dx}}{\cos^2 v} = \sec^2 v \frac{dv}{dx}$. XIII ∴ $\frac{d}{dx}(\tan x)$ $= \sec^2 v \frac{dv}{dx}$.

51. Differentiation of $\cot v$.

 Let $y$ $= cotv$. By 26, p. 2 [§ 1], this may be written $y$ $= \frac{1}{\tan v}$. Differentiating by formula VII, $\frac{dy}{dx}$ $= - \frac{\frac{d}{dx}(\tan v)}{\tan^2 v}$ $= -\frac{\sec^2 \frac{dv}{dx}}{\tan^2 v} = -\csc^2 v \frac{dv}{dx}$. XIV ∴ $\frac{d}{dx}(\cot v)$ $= -\csc^2 v \frac{dv}{dx}$.

52. Differentiation of $\sec v$.

 Let $y$ $\sec v$ By 26, p. 2 [§ 1], this may be written $y$ $= \frac{1}{\cos v}$. Differentiating by formula VII, $\frac{dy}{dx}$ $= -\frac{\frac{d}{dx}(\cos v)}{\cos^2 v}$ $=\frac{\sin v \frac{dv}{dx}}{\cos^2 v}$ $= \frac{1}{\cos v} \frac{\sin v}{\cos v} \frac{dv}{dx}$ $= \sec v \tan v \frac{dv}{dx}$. XV ∴ $\frac{d}{dx}(\sec v)$ $= \sec v \tan v \frac{dv}{dx}$.

53. Differentiation of $\csc v$.

 Let $y$ $= \csc v$. By 26, p. 2 [§ 1], this may be written $y$ $= \frac{1}{\sin v}$. Differentiating by formula VII, $\frac{dy}{dx}$ $= -\frac{\frac{d}{dx}(\sin v)}{\sin^2 v}$ $= -\frac{\cos v \frac{dv}{dx}}{\sin^2 v}$ $= -\csc v \cot v \frac{dv}{dx}$. XVI ∴ $\frac{d}{dx}(\csc v)$ $= - \csc v \cot v \frac{dv}{dx}$.

1. The student must not forget that this function is defined only for positive values of the base $a$ and the variable $v$.
2. If we take the third and fourth steps without transforming the right-hand member, there results:
Third step: $\frac{\Delta y}{\Delta v} = \frac{\log_a(v + \Delta v) - \log_a v}{\Delta v}$.
Fourth step. $\frac{dy}{dx} = \frac{0}{0}$, which is indeterminate. Hence the limiting value of the right-hand dv 0 member in the third step cannot be found by direct substitution, and the above transformation is necessary.[/itex]
3. The logarithm of $e$ to any base $a (= \log_a e)$ is called the modulus of the system whose base is $a$. In Algebra it is shown that we may find the logarithm of a number $N$ to any base $a$ by means of the formula
$\log_a N = \log_a e \cdot \log_e N = \frac{\log_e N}{\log_e a}$.
The modulus of the common or Briggs system with base 10 is
$\log_{10} e = .434294...$.
4. $u$ can here assume only positive values.
5. If we take the third and fourth steps without transforming the right-hand member, there results:
Third step. $\frac{\Delta y}{\Delta v} = \frac{\sin(v + \Delta v) - \sin v}{\Delta v}$
Fourth step. $\frac{dy}{dv} = \frac{0}{0}$, which is indeterminate (see footnote, p. 46 [§ 44]).
6.  Let $A$ $= v + \Delta v$ $A$ $= v + \Delta v$ and $B$ $= v$ $B$ $= v$ Adding, $A + B$ $= 2v + \Delta v$. Subtracting, $A - B$ $= \Delta v$ Therefore $\frac{1}{2}(A + B)$ $= v + \frac{\Delta v}{2}$. $\frac{1}{2}(A - B)$ $= \frac{\Delta v}{2}$.

Substituting these values of $A, B, \frac{1}{2}(A + B), \frac{1}{2}(A - B)$ in terms of $v$ and $\Delta v$ in the formula from Trigonometry (42, p. 2 [§ 1]),

 $\sin A - \sin B$ $= 2 \cos \frac{1}{2} (A + B) \sin \frac{1}{2} (A - B)$, we get $sin(v + \Delta v) - \sin v$ $= 2 \cos \left ( v + \frac{\Delta v}{2} \right ) sin \frac{\Delta v}{2}$.