Elements of the Differential and Integral Calculus/Chapter V

by William Anthony Granville

CHAPTER V

RULES FOR DIFFERENTIATING STANDARD ELEMENTARY FORMS

33. Importance of General Rule. The General Rule for differentiation, given in the last chapter, p. 29 [§ 31], is fundamental, being found directly from the definition of a derivative, and it is very important that the student should be thoroughly familiar with it. However, the process of applying the rule to examples in general has been found too tedious or difficult; consequently special rules have been derived from the General Rule for differentiating certain standard forms of frequent occurrence in order to facilitate the work.

It has been found convenient to express these special rules by means of formulas, a list of which follows. The student should not only memorize each formula when deduced, but should be able to state the corresponding rule in words. In these formulas $u,v$, and $w$ denote variable quantities which are functions of $x$, and are differentiable.

FORMULAS FOR DIFFERENTIATION

I		${\frac {dc}{dx}}$	$=0$.
II		${\frac {dx}{dx}}$	$=1$.
III		${\frac {d}{dx}}(u+v-w)$	$={\frac {du}{dx}}\ +\ {\frac {dv}{dx}}\ -\ {\frac {dw}{dx}}$.
IV		${\frac {d}{dx}}(cv)$	$=c{\frac {dv}{dx}}$.
V		${\frac {d}{dx}}(uv)$	$=u{\frac {dv}{dx}}\ +\ v{\frac {du}{dx}}$.
VI		${\frac {d}{dx}}\left(v^{n}\right)$	$=nv^{n-1}{\frac {dv}{dx}}$.
VI	a	${\frac {d}{dx}}\left(x^{n}\right)$	$=nx^{n-1}$.
VII		${\frac {d}{dx}}\left({\frac {u}{v}}\right)$	$={\frac {v{\frac {du}{dx}}\ -\ u{\frac {dv}{dx}}}{v^{2}}}$.
VII	a	${\frac {d}{dx}}\left({\frac {u}{c}}\right)$	$={\frac {\frac {du}{dx}}{c}}$.
VIII		${\frac {d}{dx}}\left(\log _{a}v\right)$	$=\log _{a}\ e\ \cdot \ {\frac {\frac {dv}{dx}}{v}}$.
IX		${\frac {d}{dx}}\left(a^{v}\right)$	$=a^{v}\ \log \ a\ {\frac {dv}{dx}}$.
IX	a	${\frac {d}{dx}}\left(e^{v}\right)$	$=e^{v}{\frac {dv}{dx}}$.
X		${\frac {d}{dx}}\left(u^{v}\right)$	$=vu^{v-1}{\frac {du}{dx}}\ +\ \log \ u\ \cdot \ u^{v}{\frac {dv}{dx}}$.
XI		${\frac {d}{dx}}(\sin \ v)$	$=\cos \ v{\frac {dv}{dx}}$.
XII		${\frac {d}{dx}}(\cos \ v)$	$=-\sin \ v{\frac {dv}{dx}}$.
XIII		${\frac {d}{dx}}(\tan \ v)$	$\sec ^{2}v{\frac {dv}{dx}}$.
XIV		${\frac {d}{dx}}(\cot \ x)$	$=-\csc ^{2}v{\frac {dv}{dx}}$.
XV		${\frac {d}{dx}}(\sec \ v)$	$=\sec \ v\ \tan \ v{\frac {dv}{dx}}$.
XVI		${\frac {d}{dx}}(\csc \ v)$	$=-\csc \ v\ \cot \ v{\frac {dv}{dx}}$.
XVII		${\frac {d}{dx}}(\operatorname {vers} \ v)$	$=\sin \ v{\frac {dv}{dx}}$.
XVIII		${\frac {d}{dx}}(\arcsin \ v)$	$={\frac {\frac {dv}{dx}}{\sqrt {1-v^{2}}}}$.
XIX		${\frac {d}{dx}}(\arccos \ v)$	$=-{\frac {\frac {dv}{dx}}{\sqrt {1-v^{2}}}}$.
XX		${\frac {d}{dx}}(\arctan \ v)$	$={\frac {\frac {dv}{dx}}{1+v^{2}}}$.
XXI		${\frac {d}{dx}}(\operatorname {arccot} \ v)$	$=-{\frac {\frac {dv}{dx}}{1+v^{2}}}$.
XXII		${\frac {d}{dx}}(\operatorname {arcsec} \ v)$	$={\frac {\frac {dv}{dx}}{v{\sqrt {v^{2}-1}}}}$.
XXIII		${\frac {d}{dx}}(\operatorname {arccsc} \ v)$	$=-{\frac {\frac {dv}{dx}}{v{\sqrt {v^{2}-1}}}}$.
XXIV		${\frac {d}{dx}}(\operatorname {arcvers} \ v)$	$={\frac {\frac {dv}{dx}}{\sqrt {2v-v^{2}}}}$.
XXV		${\frac {dy}{dx}}$	$={\frac {dy}{dv}}\cdot {\frac {dv}{dx}}$, $y$ being a function of $v$.
XXVI		${\frac {dy}{dx}}$	$={\frac {1}{\frac {dx}{dy}}}$, $y$ being a function of $x$.

34. Differentiation of a constant. A function that is known to have the same value for every value of the independent variable is constant, and we may denote it by

	$y$	$=c$.
As $x$ takes on an increment $\Delta x$, the function does not change in value, that is, $\Delta y=0$, and
	${\frac {\Delta y}{\Delta x}}$	$=0$
But	$\lim _{\Delta x\to 0}\left({\frac {\Delta y}{\Delta x}}\right)$	$={\frac {dy}{dx}}=0$.
I	∴ ${\frac {dc}{dx}}$'	$=0$.

The derivative of a constant is zero.

35. Differentiation of a variable with respect to itself.

Let	$y$	$=x$.
Following the General Rule, p. 29 [§ 31], we have
FIRST STEP.	$y+\Delta y$	$=x+\Delta x$.
SECOND STEP.	$\Delta y$	$=\Delta x$
THIRD STEP.	${\frac {\Delta y}{\Delta x}}$	$=1$.
FOURTH STEP.	${\frac {dy}{dx}}$	$=1$.
II	∴ ${\frac {dy}{dx}}$	$=1$.

The derivative of a variable with respect to itself is unity.

36. Differentiation of a sum.

Let	$y$	$=u+v-w$.
By the General Rule,
FIRST STEP.	$y+\Delta y$	$=u+\Delta u+v+\Delta v-w-\Delta w$.
SECOND STEP.	$\Delta y$	$=\Delta u+\Delta v-\Delta w$
THIRD STEP.	${\frac {\Delta y}{\Delta x}}$	$={\frac {\Delta u}{\Delta x}}+{\frac {\Delta v}{\Delta x}}-{\frac {\Delta w}{\Delta x}}$.
FOURTH STEP.	${\frac {dy}{dx}}$	$={\frac {du}{dx}}+{\frac {dv}{dx}}-{\frac {dw}{dx}}$.
[Applying Th. I, p. 18. (§ 20)]
III	∴ ${\frac {d}{dx}}(u+v-w)$	$={\frac {du}{dx}}+{\frac {dv}{dx}}-{\frac {dw}{dx}}$.

Similarly, for the algebraic sum of any finite number of functions.

The derivative of the algebraic sum of a finite number of functions is equal to the same algebraic sum of their derivatives.

37. Differentiation of the product of a constant and a function

Let	$y$	$=cv$.
By the General Rule,
FIRST STEP.	$y+\Delta y$	$=c(v+\Delta v)=cv+c\Delta v$.
SECOND STEP.	$\Delta y$	$=c\cdot \Delta v$
THIRD STEP.	${\frac {\Delta y}{\Delta x}}$	$=c{\frac {\Delta v}{\Delta x}}$.
FOURTH STEP.	${\frac {dy}{dx}}$	$=c{\frac {dv}{dx}}$.
[Applying Th. II, p. 18. (§ 20)]
IV	∴ ${\frac {d}{dx}}(cv)$	$=c{\frac {dv}{dx}}$.

The derivative of the product of a constant and a function is equal to the product of the constant and the derivative of the function.

38. Differentiation of the product of two functions.

Let	$y$	$=uv$.
By the General Rule,
FIRST STEP.	$y+\Delta y$	$=(u+\Delta u)(v+\Delta v)$
Multiplying out this becomes
	$y+\Delta y$	$=uv+u\cdot \Delta v+v\cdot \Delta u+\Delta u\cdot \Delta v$.
SECOND STEP.	$\Delta y$	$=u\cdot \Delta v+v\cdot \Delta u+\Delta u\cdot \Delta v$.
THIRD STEP.	${\frac {\Delta y}{\Delta x}}$	$=u{\frac {\Delta v}{\Delta x}}+v{\frac {\Delta u}{\Delta x}}+\Delta u{\frac {\Delta v}{\Delta x}}$.
FOURTH STEP.	${\frac {dy}{dx}}$	$=u{\frac {dv}{dx}}+v{\frac {du}{dx}}$.
[Applying Th. II, p. 18 (§ 20), since when $\Delta x{\dot {=}}0,\Delta u{\dot {=}}0$, and $\left(\Delta u{\frac {\Delta v}{\Delta x}}\right){\dot {=}}0$.]
V	∴ ${\frac {d}{dx}}(uv)$	$=u{\frac {dv}{dx}}+v{\frac {du}{dx}}$.

The derivative of the product of two functions is equal to the first function times the derivative of the second, plus the second function times the derivative of the first.

39. Differentiation of the product of any finite number of functions.

Now in dividing both sides of V by $uv$, this formula assumes the form

${\frac {{\frac {d}{dx}}(uv)}{uv}}$	$={\frac {\frac {du}{dx}}{u}}+{\frac {\frac {dv}{dx}}{v}}$.
If then we have the product of $n$ functions
$y$	$=v_{1}v_{2}\cdots v_{n}$.
we may write
${\frac {{\frac {d}{dx}}(v_{1}v_{2}\cdots v_{n})}{v_{1}v_{2}\cdots v_{n}}}$	$={\frac {\frac {dv_{1}}{dx}}{v_{1}}}+{\frac {{\frac {d}{dx}}(v_{2}v_{3}\cdots v_{n})}{v_{2}v_{3}\cdots v_{n}}}$
	$={\frac {\frac {dv_{1}}{dx}}{v_{1}}}+{\frac {\frac {dv_{2}}{dx}}{v_{2}}}+{\frac {{\frac {d}{dx}}(v_{3}v_{4}\cdots v_{n})}{v_{3}v_{4}\cdots v_{n}}}$
	$={\frac {\frac {dv_{1}}{dx}}{v_{1}}}+{\frac {\frac {dv_{2}}{dx}}{v_{2}}}+{\frac {\frac {dv_{3}}{dx}}{v_{3}}}+\cdots +{\frac {\frac {dv_{n}}{dx}}{v_{n}}}$
${\frac {d}{dx}}(v_{1}v_{2}\cdots v_{n})$	$=(v_{2}v_{3}\cdots v_{n}){\frac {dv_{1}}{dx}}+(v_{1}v_{3}\cdots v_{n}){\frac {dv_{2}}{dx}}+\cdots$
	$+(v_{1}v_{2}\cdots v_{n-1}){\frac {dv_{n}}{dx}}$.

The derivative of the product of a finite number of functions is equal to the sum of all the products that can be formed by multiplying the derivative of each function by all the other functions.

40. Differentiation of a function with a constant exponent. If the $n$ factors in the above result are each equal to $v$, we get

	${\frac {{\frac {d}{dx}}(v^{n})}{v^{n}}}$	$=n{\frac {\frac {dv}{dx}}{v}}$.
VI	∴ ${\frac {d}{dx}}(v^{n})$	$=nv^{n-1}{\frac {dv}{dx}}$.
When $v=x$ this becomes
VIa	${\frac {d}{dx}}(x^{n})$	$=nx^{n-1}$.

We have so far proven VI only for the case when $n$ is a positive integer. In § 46, however, it will be shown that this formula holds true for any value of $n$, and we shall make use of this general result now.

The derivative of a function with a constant exponent is equal to the product of the exponent, the function with the exponent diminished by unity, and the derivative of the function.

41. Differentiation of a quotient.

Let	$y$	$={\frac {u}{v}}$	$v\neq 0$.
By the General Rule,
FIRST STEP.	$y+\Delta y$	$={\frac {u+\Delta u}{v+\Delta v}}$.
SECOND STEP.	$\Delta y$	$={\frac {u+\Delta u}{v\Delta v}}-{\frac {u}{v}}={\frac {v\cdot \Delta u-u\cdot \Delta v}{v(v+\Delta v)}}$.
THIRD STEP.	${\frac {\Delta y}{\Delta x}}$	$={\frac {v{\frac {\Delta u}{\Delta x}}-u{\frac {\Delta v}{\Delta x}}}{v(v+\Delta v)}}$.
FOURTH STEP.	${\frac {dy}{dx}}$	$={\frac {v{\frac {du}{dx}}-u{\frac {dv}{dx}}}{v^{2}}}$
[Applying Theorems II and III, p. 18. (§ 20)]
VII	∴ ${\frac {d}{dx}}\left({\frac {u}{v}}\right)$	${\frac {v{\frac {du}{dx}}-u{\frac {dv}{dx}}}{v^{2}}}$.

The derivative of a fraction is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.'

When the denominator is constant, set $v=c$ in VII, giving

VIIa	${\frac {d}{dx}}\left({\frac {u}{c}}\right)$	$={\frac {\frac {du}{dx}}{c}}$.
[Since ${\frac {dv}{dx}}={\frac {dc}{dx}}=0$.]
We may also get VIIa from IV as follows:
	${\frac {d}{dx}}\left({\frac {u}{c}}\right)$	$={\frac {1}{c}}{\frac {du}{dx}}={\frac {\frac {du}{dx}}{c}}$.

The derivative of the quotient of a function by a constant is equal to the derivative of the function divided by the constant.

All explicit algebraic functions of one independent variable may be differentiated by following the rules we have deduced so far.

EXAMPLES^[1]

Differentiate the following:

1. $y=x^{3}$.

Solution. ${\frac {dy}{dx}}={\frac {d}{dx}}(x^{3})=3x^{2}$. Ans.	By VIa
[$n=3$.]

2. $y=ax^{4}-bx^{2}$.

Solution. ${\frac {dy}{dx}}$	$={\frac {d}{dx}}(ax^{4}-bx^{2})={\frac {d}{dx}}(ax^{4})-{\frac {d}{dx}}(bx^{2})$	by III
	$=a{\frac {d}{dx}}(x^{4})-b{\frac {d}{dx}}(x^{2})$	by IV
	$=4ax^{3}-2bx$. Ans.	By Via

3. $y=x^{\frac {4}{3}}+5$.

Solution. ${\frac {dy}{dx}}$	$={\frac {d}{dx}}(x^{\frac {4}{3}})+{\frac {d}{dx}}(5)$	by III
	$={\frac {4}{3}}x^{\frac {1}{3}}$ Ans.	By Via and I

4. $y={\frac {3x^{3}}{\sqrt[{5}]{x^{2}}}}-{\frac {7x}{\sqrt[{3}]{x^{4}}}}+8{\sqrt[{7}]{x^{3}}}$

Solution. ${\frac {dy}{dx}}$	$={\frac {d}{dx}}\left(3x^{\frac {13}{5}}\right)+{\frac {d}{dx}}\left(7x^{-{\frac {1}{3}}}\right)+{\frac {d}{dx}}\left(8x^{\frac {3}{7}}\right)$.	by III
	$={\frac {39}{5}}x^{\frac {8}{5}}+{\frac {7}{3}}x^{-{\frac {4}{3}}}+{\frac {24}{7}}x^{-{\frac {4}{7}}}$. Ans.	By IV and Via

5. $y=(x^{2}-3)^{5}$.

Solution. ${\frac {dy}{dx}}$	$5(x^{2}-3)^{4}{\frac {d}{dx}}(x^{2}-3)$.	by VI
[$v=x^{2}-3$ and $n=5$.]
	$5(x^{2}-3)^{4}\cdot 2x=10x(x^{2}-3)^{4}$. Ans.

We might have expanded this function by the Binomial Theorem and then applied III, etc., but the above process is to be preferred.

6. $y={\sqrt {(}}a^{2}-x^{2})$.

Solution. ${\frac {dy}{dx}}$	${\frac {d}{dx}}(a^{2}-x^{2})^{\frac {1}{2}}={\frac {1}{2}}(a^{2}-x^{2})^{-{\frac {1}{2}}}{\frac {d}{dx}}(a^{2}-x^{2})$.	by VI
[$v=a^{2}-x^{2}$, and $n=5$.]
	$={\frac {1}{2}}(a^{2}-x^{2})^{-{\frac {1}{2}}}(-2x)=-{\frac {x}{\sqrt {a^{2}-x^{2}}}}$. Ans.

7. $y=(3x^{2}+2){\sqrt {1+5x^{2}}}$.

Solution. ${\frac {dy}{dx}}$	$=(3x^{2}+2){\frac {d}{dx}}(1+5x^{2})^{\frac {1}{2}}+(1+5x^{2})^{\frac {1}{2}}{\frac {d}{dx}}(3x^{2}+2)$	by V
[$u=3x^{2}+2$, and $v=(1+5x^{2})^{\frac {1}{2}}$.]
	$=(3x^{2}+2){\frac {1}{2}}(1+5x^{2})^{-{\frac {1}{2}}}{\frac {d}{dx}}(1+5x^{2})+(1+5x^{2})^{\frac {1}{2}}6x$.	by VI, etc.
	$=(3x^{2}+2)(1+5x^{2})^{-{\frac {1}{2}}}5x+6x(1+5x^{2})^{\frac {1}{2}}$
	$={\frac {5x(3x^{2}+2)}{\sqrt {1+5x^{2}}}}+6x{\sqrt {1+5x^{2}}}={\frac {45x^{3}+16x}{\sqrt {1+5x^{2}}}}$. Ans.

8. $y={\frac {a^{2}+x^{2}}{\sqrt {a^{2}-x^{2}}}}$

Solution. ${\frac {dy}{dx}}$	$={\frac {(a^{2}-x^{2})^{\frac {1}{2}}{\frac {d}{dx}}(a^{2}-x^{2})-(a^{2}+x^{2}){\frac {d}{dx}}(a^{2}-x^{2})^{\frac {1}{2}}}{a^{2}-x^{2}}}$	by VII
	$={\frac {2x(a^{2}-x^{2})+x(a^{2}+x^{2})}{(a^{2}-x^{2})^{-{\frac {3}{2}}}}}$
[Multiplying both numerator and denominator by (a^2 - x^2)^{\frac{1}{2}}${\displaystyle }$.]
	$={\frac {{\frac {3}{x}}^{2}x-x^{3}}{(a^{2}-x^{2})^{\frac {3}{2}}}}$. Ans.

9. $5x^{4}+3x^{2}-6$.	${\frac {dy}{dx}}=2x^{3}+6x$
10. $y=3cx^{2}-8dx+5e$.	${\frac {dy}{dx}}=6cx-8d$.
11. $y=x^{a+b}$.	${\frac {dy}{dx}}=(a+b)x^{a+b-1}$.
12. $y=x^{n}+nx+n$.	${\frac {dy}{dx}}=nx^{n-1}+n$.
13. $f(x)={\frac {2}{3}}x^{3}-{\frac {3}{2}}x^{2}+5$	$f'(x)=2x^{2}-3x$.
14. $f(x)=(a+b)x^{2}+cx+d$.	$f'(x)=2(a+b)x+c$.

15. ${\frac {d}{dx}}(a+bx+cx^{2})=b+2cx$.

16. ${\frac {d}{dy}}(5y^{m}-3y+6)=5my^{m-1}-3$.

17. ${\frac {d}{dx}}(2x^{-2}+3x^{-3})=-4x^{-3}-9x^{-4}$.

18. ${\frac {d}{ds}}(3s^{-4}-s)=-12s^{-5}-1$.

19. ${\frac {d}{dx}}(4x^{\frac {1}{2}}+x^{2})=2x^{-{\frac {1}{2}}}+2x$.

20. ${\frac {d}{dy}}(y^{-2}-4y^{-{\frac {1}{2}}}=-2y^{-3}+2y^{-{\frac {3}{2}}}$.

21. ${\frac {d}{dx}}(2x^{3}+5)=6x^{2}$.

22. ${\frac {d}{dt}}(3t^{5}-2t^{2})=15t^{4}-4t$.

23. ${\frac {d}{d\theta }}(a\theta ^{4}+b\theta )=4a\theta ^{3}+b$

24. ${\frac {d}{d\alpha }}(5-2\alpha ^{\frac {3}{2}})=-3\alpha ^{\frac {1}{2}}$

25. ${\frac {d}{dt}}(9t^{\frac {5}{3}}+t^{-1})=15t^{\frac {2}{3}}-t^{-2}$.

26. ${\frac {d}{dx}}(2x^{12}-x^{9})=24x^{11}-9x^{8}$.

27. $r=c\theta ^{3}+d\theta ^{2}+e\theta$.	$r'=3c\theta ^{2}+2d\theta +e$.
28. $y=6x^{\frac {7}{2}}+4x^{\frac {5}{2}}+2x^{\frac {3}{2}}$.	$y'=21x^{\frac {5}{2}}+10x^{\frac {3}{2}}+3x^{\frac {1}{2}}$.
29. $y={\sqrt {3x}}+{\sqrt {3}}{x}+{\frac {1}{x}}$.	$y'={\frac {3}{2{\sqrt {3x}}}}+{\frac {1}{3{\sqrt[{3}]{x^{2}}}}}-{\frac {1}{x^{2}}}$.
30. $y={\frac {a+bx+cx^{2}}{x}}$.	$y'=c-{\frac {a}{x^{2}}}$.
31. $y={\frac {(x-1)^{3}}{x^{\frac {1}{3}}}}$	$y'={\frac {8}{3}}x^{\frac {5}{3}}-5x^{\frac {2}{3}}+2x^{-{\frac {1}{3}}}+{\frac {1}{3}}x^{-{\frac {4}{3}}}$.
32. $y={\frac {x^{\frac {5}{2}}-x-x^{\frac {1}{2}}+a}{x^{\frac {3}{2}}}}$.	$y'={\frac {2x^{\frac {5}{2}}+x+2x^{\frac {1}{2}}-3a}{2x^{\frac {5}{2}}}}$.
33. $y=(2x^{3}+x^{2}-5)^{3}$.	$y'=6x(3x+1)(2x^{3}+x^{2}-5)^{2}$.
34. $f(x)=(a+bx^{2})^{\frac {5}{4}}$.	$f'(x)={\frac {5bx}{2}}(a+bx^{2})^{\frac {1}{4}}$.
35. $f(x)=(1+4x^{3})(1+2x^{2})$.	$f'(x)=4x(1+3x+10x^{3})$.
36. $f(x)=(a+x){\sqrt {a-x}}$.	$f'(x)={\frac {a-3x}{2{\sqrt {a-x}}}}$.
37. $f(x)=(a+x)^{m}(b+x)^{n}$.	$f'(x)=(a+x)^{m}(b+x)^{n}\left[{\frac {m}{a+x}}+{\frac {n}{b+x}}\right]$.
38. $y={\frac {1}{x^{n}}}$	${\frac {y}{x}}=-{\frac {n}{x^{n+1}}}$
39. $y=x(a^{2}+x^{2}){\sqrt {a^{2}-x^{2}}}$.	${\frac {dy}{dx}}={\frac {a^{4}+a^{2}x^{2}-4x^{4}}{\sqrt {a^{2}-x^{2}}}}$.

40. Differentiate the following functions:

(a) ${\frac {d}{dx}}(2x^{3}-4x+6)$.	(e) ${\frac {d}{dt}}(b+at^{2})^{\frac {1}{2}}$.	(i) ${\frac {d}{dx}}(x^{\frac {2}{3}}-a^{\frac {2}{3}}$.
(b) ${\frac {d}{dt}}(at^{7}+bt^{5}-9)$.	(f) ${\frac {d}{dx}}(x^{2}-a^{2})^{\frac {3}{2}}$.	(j) ${\frac {d}{dt}}(5+2t)^{\frac {9}{2}}$
(c) ${\frac {d}{d\theta }}(3\theta ^{\frac {3}{2}}-2\theta ^{\frac {1}{2}}+6\theta )$.	(g) ${\frac {d}{d\phi }}(4-\phi ^{\frac {2}{5}})$.	(k) ${\frac {d}{ds}}{\sqrt {a+b{\sqrt {s}}}}$.
(d) ${\frac {d}{dx}}(2x^{3}+x)^{\frac {5}{3}}$.	(h) ${\frac {d}{dt}}{\sqrt {1+9t^{2}}}$.	(l) ${\frac {d}{dx}}(2x^{\frac {1}{3}}+2x^{\frac {5}{3}})$.

41. $y={\frac {2x^{4}}{b^{2}-x^{2}}}$.	${\frac {dy}{dx}}={\frac {8b^{2}x^{3}-4x^{5}}{(b^{2}-x^{2})^{2}}}$.
42. $y={\frac {a-x}{a+x}}$	${\frac {dy}{dx}}=-{\frac {2a}{(a+x)^{2}}}$
43. $s={\frac {t^{3}}{(1+t)^{2}}}$.	${\frac {ds}{dt}}={\frac {3t^{2}+t^{3}}{(1+t)^{3}}}$.
44. $f(s)={\frac {(s+4)^{2}}{s+3}}$	$f'(s)={\frac {(s+2)(s+4)}{(s+3)^{2}}}$.
45. $f(\theta )={\frac {\theta }{\sqrt {a-b\theta ^{2}}}}$.	$f'(\theta )={\frac {a}{(a-b\theta ^{2})^{\frac {3}{2}}}}$.
46. $F(r)={\sqrt {\frac {1+r}{1-r}}}$	$F'(r)={\sqrt {1}}{(1-r){\sqrt {1-r^{2}}}}$.
47. $\psi (y)=\left({\frac {y}{1-y}}\right)^{m}$.	$\psi '(y)={\frac {my^{m-1}}{(1-y)^{m+1}}}$.
48. $\phi (x)={\frac {2x^{2}-1}{x{\sqrt {1+x^{2}}}}}$.	$\phi '(x)={\frac {1+4x^{2}}{x^{2}(1+x^{2})^{\frac {3}{2}}}}$.
49. $y={\sqrt {2px}}$.	$y'={\frac {p}{y}}$.
50. $y={\frac {b}{a}}{\sqrt {a^{2}-x^{2}}}$.	$y'=-{\frac {b^{2}x}{a^{2}y}}$.
51. $y=(a^{\frac {2}{3}}-x^{\frac {2}{3}})^{\frac {3}{2}}$.	$y'=-{\sqrt[{3}]{\frac {y}{x}}}$.
52. $r={\sqrt {a\phi }}+c{\sqrt {\phi ^{3}}}$.	$r'={\frac {{\sqrt {a}}+3c\phi }{2{\sqrt {\phi }}}}$.
53. $u={\frac {v^{c}+v^{d}}{cd}}$.	$u'={\frac {v^{c-1}}{d}}+{\frac {v^{d-1}}{c}}$.
54. $p={\frac {(q+1)^{\frac {3}{2}}}{\sqrt {q-1}}}$.	$p'={\frac {(q-2){\sqrt {q+1}}}{(q-1)^{\frac {3}{2}}}}$.

55. Differentiate the following functions:

(a) ${\frac {d}{dx}}\left({\frac {a^{2}-x^{2}}{a^{2}+x^{2}}}\right)$.	(d) ${\frac {d}{dy}}\left({\frac {ay^{2}}{b+y^{3}}}\right)$.	(g) ${\frac {d}{dx}}{\frac {x^{2}}{\sqrt {1-x^{2}}}}$.
(b) ${\frac {d}{dx}}\left({\frac {x^{3}}{1+x^{4}}}\right)$.	(e) ${\frac {d}{ds}}\left({\frac {a^{2}-s^{2}}{\sqrt {a^{2}+s^{2}}}}\right)$.	(h) ${\frac {d}{dx}}{\frac {1+x^{2}}{(1-x^{2})^{\frac {3}{2}}}}$.
(c) ${\frac {d}{dx}}\left({\frac {1+x}{\sqrt {1-x}}}\right)$.	(f) ${\frac {d}{dx}}{\frac {\sqrt {4-2x^{3}}}{x}}$.	(i) ${\frac {d}{dt}}{\sqrt {\frac {1+t^{2}}{1-t^{2}}}}$.

42. Differentiation of a function of a function. It sometimes happens that $y$, instead of being defined directly as a function of $x$, is given as a function of another variable $v$, which is defined as a function of $x$. In that case $y$ is a function of $x$ through $v$ and is called a function of a function.

For example, if	$y={\frac {2v}{1-v^{2}}}$,
and	$v=1-x^{2}$,

then $y$ is a function of a function. By eliminating $v$ we may express $y$ directly as a function of $x$, but in general this is not the best plan when we wish to find ${\frac {dy}{dx}}$.

If $y=f(v)$ and $v=\phi (x)$, then $y$ is a function of $x$ through $v$. Hence, when we let $x$ take on an increment $\Delta x$, $v$ will take on an increment $\Delta v$ and $y$ will also take on a corresponding increment $\Delta y$. Keeping this in mind, let us apply the General Rule simultaneously to the two functions $y=f(v)$ and $v=\phi (x)$.

FIRST STEP.	$y+\Delta y$	$=f(v+\Delta v)$	$v+\Delta v$	$=\phi (x+\Delta x)$
SECOND STEP	$y+\Delta y$	$=f(v+\Delta v)$	$v+\Delta v$	$=\phi (x+\Delta x)$
	$y$	$=f(v)$	$v$	$=\phi (x)$
	$\Delta y$	$f(v+\Delta v)-f(v)$,	$\Delta v$	$\phi (x+\Delta x)-\phi (x)$
THIRD STEP.	${\frac {\Delta y}{\Delta v}}$	$={\frac {f(v+\Delta v)-f(v)}{\Delta v}}$	${\frac {\Delta v}{\Delta x}}$	${\frac {\phi (x+\Delta x)-\phi (x)}{\Delta x}}$

The left-hand members show one form of the ratio of the increment of each function to the increment of the corresponding variable, and the right-hand members exhibit the same ratios in another form. Before passing to the limit let us form a product of these two ratios, choosing the left-hand forms for this purpose.

This gives ${\frac {\Delta y}{\Delta v}}\cdot {\frac {\Delta v}{\Delta x}}$, which equals ${\frac {\Delta y}{\Delta x}}$.

Write this	${\frac {\Delta y}{\Delta x}}$	$={\frac {\Delta y}{\Delta v}}\cdot {\frac {\Delta v}{\Delta x}}$.
FOURTH STEP. Passing to the limit,
(A)	${\frac {dy}{dx}}$	$={\frac {dy}{dv}}\cdot {\frac {dv}{dx}}$.	Th. II, p. 18 [§20]
This may also be written
(B)	${\frac {dy}{dx}}$	$=f'(v)\cdot \phi '(x)$.

If $y=f(v)$ and $v=\phi (x)$, the derivative of $y$ with respect to $x$ equals the product of the derivative of $y$ with respect to $v$ and the derivative of $v$ with respect to $x$.

43. Differentiation of inverse functions. Let $y$ be given as a function of $x$ by means of the relation $y=f(x)$.

It is usually possible in the case of functions considered in this book to solve this equation for $x$, giving

$x=\phi (y)$;

that is, to consider $y$ as the independent and $x$ as the dependent variable. In that case $f(x)$ and $\phi (y)$ are said to be inverse functions. When we wish to distinguish between the two it is customary to call the first one given the direct function and the second one the inverse function. Thus, in the examples which follow, if the second members in the first column are taken as the direct functions, then the corresponding members in the second column will be respectively their inverse functions.

$y=x^{2}+1$,	$x=\pm {\sqrt {y-1}}$.
$y=a^{x}$,	$x=\log _{a}y$.
$y=\sin \ x$,	$x=\arcsin \ y$.

Let us now differentiate the inverse functions

$y=f(x)$ and $x=\phi (y)$

simultaneously by the General Rule.

FIRST STEP.	$y+\Delta y$	$=f(x+\Delta x)$	$x+\Delta x$	$=\phi (y+\Delta y)$
SECOND STEP.	$y+\Delta y$	$=f(x+\Delta x)$	$x+\Delta x$	$=\phi (y+\Delta y)$
	$y$	$=f(x)$	$x$	$=\phi (y)$
$\Delta y$	$=f(x+\Delta x)-f(x)$	$\Delta x$	$=\phi (y+\Delta y)-\phi (y)$
THIRD STEP.	${\frac {\Delta y}{\Delta x}}$	$={\frac {f(x+\Delta x)-f(x)}{\Delta x}}$	${\frac {\Delta x}{\Delta y}}$	$={\frac {\phi (y+\Delta y)-\phi (y)}{\Delta y}}$

Taking the product of the left-hand forms of these ratios, we get

	${\frac {\Delta y}{\Delta x}}\cdot {\frac {\Delta x}{\Delta y}}$	$=1$.
or,	${\frac {\Delta y}{\Delta x}}$	$={\frac {1}{\frac {\Delta x}{\Delta y}}}$.

FOURTH STEP. Passing to the limit,

(C)	${\frac {dy}{dx}}$	$={\frac {1}{\frac {dx}{dy}}}$,
or,
(D)	$f'(x)$	$={\frac {1}{\phi '(y)}}$.

The derivative of the inverse function is equal to the reciprocal of the derivative of the direct function.

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1. When learning to differentiate, the student should have oral drill in differentiating simple functions.