Elements of the Differential and Integral Calculus/Chapter VII

by William Anthony Granville

CHAPTER VII

SUCCESSIVE DIFFERENTIATION

74. Definition of successive derivatives. We have seen that the derivative of a function of x is in general also a function of x. This new function may also be differentiable, in which case the derivative of the first derivative is called the second derivative of the original function. Similarly, the derivative of the second derivative is called the third derivative; and so on to the nth derivative. Thus, if

$\ y$	$=\ 3x^{4}$
${\frac {dy}{dx}}$	$=\ 12x^{3}$
${\frac {d}{dx}}\left({\frac {dy}{dx}}\right)$	$=\ 36x^{2}$
${\frac {d}{dx}}\left[{\frac {d}{dx}}\left({\frac {dy}{dx}}\right)\right]$	$=\ 72x$, etc.

75. Notation. The symbols for the successive derivatives are usually abbreviated as follows:

${\frac {d}{dx}}\left({\frac {dy}{dx}}\right)$	$=\ {\frac {d^{2}y}{dx^{2}}}$
${\frac {d}{dx}}\left[{\frac {d}{dx}}\left({\frac {dy}{dx}}\right)\right]$	$=\ {\frac {d}{dx}}\left({\frac {d^{2}y}{dx^{2}}}\right)={\frac {d^{3}y}{dx^{3}}}$,
. . .	. . .
${\frac {d}{dx}}\left({\frac {d^{n-1}y}{dx^{n-1}}}\right)$	$={\frac {d^{n}y}{dx^{n}}}$.

If $y=f(x)$, the successive derivatives are also denoted by

$f'(x),\ f''(x),\ f'''(x),\ f^{iv}(x),\ ...,\ f^{(n)}(x)$;

$y',\ y'',\ y''',\ y^{iv},\ ...,\ y^{(n)}$;

or,

${\frac {d}{dx}}f(x),\ {\frac {d^{2}}{dx^{2}}}f(x),\ {\frac {d^{3}}{dx^{3}}}f(x),\ {\frac {d^{4}}{dx^{4}}}f(x)\ ...,\ {\frac {d^{n}}{dx^{n}}}f(x)$.

76. The nth derivative. For certain functions a general expression involving n may be found for the nth derivative. The usual plan is to find a number of the first successive derivatives, as many as may be necessary to discover their law of formation, and then by induction write down the nth derivative.

Illustrative Example 1. Given $y=e^{ax}$, find ${\frac {d^{n}y}{dx^{n}}}$.

Solution,	${\frac {dy}{dx}}$	$=ae^{ax}$,
	${\frac {d^{2}y}{dx^{2}}}$	$=a^{2}e^{ax}$
	.   .   .
∴	${\frac {d^{n}y}{dx^{n}}}$	$=a^{n}e^{ax}$. Ans.

Illustrative Example 2. Given $y=\ln x$, find ${\frac {d^{n}y}{dx^{n}}}$

Solution,	${\frac {dy}{dx}}$	$={\frac {1}{x}}$,
	${\frac {d^{2}y}{dx^{2}}}$	$=-{\frac {1}{x^{2}}}$
	${\frac {d^{3}y}{dx^{3}}}$	$={\frac {1\cdot 2}{x^{3}}}$,
	${\frac {d^{4}y}{dx^{4}}}$	$={\frac {1\cdot 2\cdot 3}{x^{4}}}$,
	.   .   .
∴	${\frac {d^{n}y}{dx^{n}}}$	$=(-1)^{n-1}{\frac {(n-1)!}{x^{n}}}$. Ans.

Illustrative Example 3. Given $y=\sin x$, find ${\frac {d^{n}y}{dx^{n}}}$

Solution,	${\frac {dy}{dx}}=\cos x$	$=\sin \left(x+{\frac {\pi }{2}}\right)$,
	${\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\sin \left(x+{\frac {\pi }{2}}\right)$	$=\cos \left(x+{\frac {\pi }{2}}\right)=\sin \left(x+{\frac {2\pi }{2}}\right)$,
	${\frac {d^{3}y}{dx^{3}}}={\frac {d}{dx}}\sin \left(x+{\frac {2\pi }{2}}\right)$	$=\cos \left(x+{\frac {2\pi }{2}}\right)=\sin \left(x+{\frac {3\pi }{2}}\right)$
	.   .   .	.   .   .
∴	${\frac {d^{n}y}{dx^{n}}}$	$=\sin \left(x+{\frac {n\pi }{2}}\right)$. Ans.

77. Leibnitz's Formula for the nth derivative of a product. This formula expresses the nth derivative of the product of two variables in terms of the variables themselves and their successive derivatives.

If u and v are functions of x, we have, from V, §33,

${\frac {d}{dx}}(uv)={\frac {du}{dx}}v+u{\frac {dv}{dx}}$.

Differentiating again with respect to x,

${\frac {d^{2}}{dx^{2}}}(uv)={\frac {d^{2}u}{dx^{2}}}v+{\frac {du}{dx}}{\frac {dv}{dx}}+{\frac {du}{dx}}{\frac {dv}{dx}}+u{\frac {d^{2}v}{dx^{2}}}={\frac {d^{2}u}{dx^{2}}}v+2{\frac {du}{dx}}{\frac {dv}{dx}}+u{\frac {d^{2}v}{dx^{2}}}$

Similarly,

${\frac {d^{3}}{dx^{3}}}(uv)$	$={\frac {d^{3}u}{dx^{3}}}v+{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+2{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+2{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+u{\frac {d^{3}v}{dx^{3}}}$
	$={\frac {d^{3}u}{dx^{3}}}v+3{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+3{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+u{\frac {d^{3}v}{dx^{3}}}$.

However far this process may be continued, it will be seen that the numerical coefficients follow the same law as those of the Binomial Theorem, and the indices of the derivatives correspond to the exponents of the Binomial Theorem.^[1] Reasoning them by mathematical induction from the mth to the (m + 1 )th derivative of the product, we can prove Leibnitz's Formula

${\frac {d^{n}}{dx^{n}}}(uv)={\frac {d^{n}u}{dx^{n}}}v+n{\frac {d^{n-1}u}{dx^{n-1}}}{\frac {dv}{dx}}+{\frac {n(n-1)}{2!}}{\frac {d^{n-2}u}{dx^{n-2}}}{\frac {d^{2}v}{dx^{2}}}+\cdots +$
$n{\frac {du}{dx}}{\frac {d^{n-1}v}{dx^{n-1}}}+u{\frac {d^{n}}{dx^{n}}}v$.

Illustrative Example 1. Given $y=e^{x}\ln x$, find ${\frac {d^{3}y}{dx^{3}}}$ by Leibnitz's Formula.

Solution. Let	$u=e^{x}$, and	$v=\log x$;
then	${\frac {du}{dx}}=e^{x}$,	${\frac {dv}{dx}}={\frac {1}{x}}$,
	${\frac {d^{2}u}{dx^{2}}}=e^{x}$,	${\frac {d^{2}v}{dx^{2}}}=-{\frac {1}{x^{2}}}$,
	${\frac {d^{3}u}{dx^{3}}}=e^{x}$	${\frac {d^{3}v}{dx^{3}}}={\frac {2}{x^{3}}}$.

Substituting in (17), we get

${\frac {d^{3}y}{dx^{3}}}=e^{x}\log x+{\frac {3e^{x}}{x}}-{\frac {3e^{x}}{x^{2}}}=e^{x}\left(\log x+{\frac {3}{x}}-{\frac {3}{x^{2}}}+{\frac {2}{x^{3}}}\right)$.

Illustrative Example 2. Given $y=x^{2}e^{ax}$, find ${\tfrac {d^{n}y}{dx^{n}}}$ by Leibnitz's Formula.

Solution. Let	$u=x^{2}$,	and	$v=e^{ax}$;
then	${\frac {du}{dx}}=2x$	${\frac {dv}{dx}}=ae^{ax}$,
	${\frac {d^{2}u}{dx^{2}}}=2$,	${\frac {d^{2}v}{dx^{2}}}=a^{2}e^{ax}$,
	${\frac {d^{3}u}{dx^{3}}}=0$,	${\frac {d^{3}v}{dx^{3}}}=a^{3}e^{ax}$,
	.   .   .	.   .   .
	${\frac {d^{n}u}{dx^{n}}}=0$,	${\frac {d^{n}v}{dx^{n}}}=a^{n}e^{ax}$.

Substituting in (17), we get

${\frac {d^{n}y}{dx^{n}}}=x^{2}a^{n}e^{ax}+2na^{n-1}xe^{ax}+n(n-1)a^{n-2}e^{ax}=a^{n-2}e^{ax}[x^{2}a^{2}+2nax+n(n-1)]$.

78. Successive differentiation of implicit functions. To illustrate the process we shall find ${\frac {d^{2}y}{dx^{2}}}$ from the equation of the hyperbola

$b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2}$.

Differentiating with respect to x, as in § 63,

$2b^{2}x-2a^{2}y{\frac {dy}{dx}}=0$,

or,

(A)	${\frac {dy}{dx}}={\frac {b^{2}x}{a^{2}y}}$.

Differentiating again, remembering that y is a function of x,

${\frac {d^{2}y}{dx^{2}}}={\frac {a^{2}yb^{2}-b^{2}xa^{2}{\frac {dy}{dx}}}{a^{4}y^{2}}}$

Substituting for ${\frac {dy}{dx}}$ its value from (A),

${\frac {d^{2}y}{dx^{2}}}={\frac {a^{2}b^{2}y-a^{2}b^{2}x\left({\frac {b^{2}y}{a^{2}y}}\right)}{a^{4}y^{2}}}=-{\frac {b^{2}(b^{2}x^{2}-a^{2}y^{2})}{a^{4}y^{3}}}$.

But from the given equation, $b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2}$.

∴ ${\frac {d^{2}y}{dx^{2}}}=-{\frac {b^{4}}{a^{2}y^{3}}}$.

EXAMPLES

Differentiate the following:

1. $y=4x^{3}-6x^{2}+4x+7$.	${\frac {d^{2}y}{dx^{2}}}=12(2x-1)$.
2. $f(x)={\frac {x^{3}}{1-x}}$.	$f^{iv}(x)={\frac {4!}{(1-x)^{5}}}$.
3. $f(y)=y^{6}$.	$f^{vi}(y)=6!$.
4. $y=x^{3}\log x$.	${\frac {d^{4}y}{dx^{4}}}={\frac {6}{x}}$.
5. $y={\frac {c}{x^{n}}}$.	$y''={\frac {n(n+1)c}{x^{n+2}}}$.
6. $y=(x-3)e^{2x}+4xe^{x}+x$.	$y''=4e^{x}[(x-2)e^{x}+x+2]$.
7. $y={\frac {a}{2}}(e^{\frac {x}{a}}+e^{-{\frac {x}{a}}})$.	$y''={\frac {1}{2a}}(e^{\frac {x}{a}}+e^{-{\frac {x}{a}}})={\frac {y}{a^{2}}}$.
8. $f(x)=ax^{2}+bx+c$.	$f'''(x)=0$.
9. $f(x)=\log(x+1)$.	$f^{iv}(x)=-{\frac {6}{(x+1)^{4}}}$.
10. $f(x)=\log(e^{x}+e^{-x})$.	$f'''(x)=-{\frac {8(e^{x}-e^{-x})}{(e^{x}-e^{-x})^{3}}}$.
11. $r=\sin a\theta$.	${\frac {d^{4}r}{d\theta ^{4}}}=a^{4}\sin a\theta =a^{4}r$.
12. $r=\tan \phi$.	${\frac {d^{3}r}{d\phi ^{3}}}=6\sec ^{6}\phi -4\sec ^{2}\phi$.
13. $r=\log \sin \phi$	$r'''=2\cot \phi \csc ^{2}\phi$.
14. $f(t)=e^{-t}\cos t$.	$f^{iv}(t)=-4e^{-t}\cos t=-4f(t)$.
15. $f(\theta )={\sqrt {\sec 2\theta }}$.	$f''(\theta )=3[f(\theta )]^{5}-f(\theta )$.
16. $p=(q^{2}+a^{2})\arctan {\frac {q}{a}}$.	${\frac {d^{3}p}{dq^{3}}}={\frac {4a^{3}}{(a^{2}+q^{2})^{2}}}$.
17. $y=a^{x}$.	${\frac {d^{n}y}{dx^{n}}}=(\log a)^{n}a^{x}$.
18. $y=\log(1+x)$.	${\frac {d^{n}y}{dx^{n}}}=(-1)^{n-1}{\frac {(n-1)!}{(1+x)^{n}}}$.
19. $y=\cos ax$	${\frac {d^{n}y}{dx^{n}}}=a^{n}\cos \left(ax+{\frac {n\pi }{2}}\right)$.
20. $y=x^{n-1}\log x$	${\frac {d^{n}y}{dx^{n}}}={\frac {(n-1)!}{x}}$.
[n = a positive integer]
21. $y={\frac {1-x}{1+x}}$.	${\frac {d^{n}y}{dx^{n}}}=2(-1)^{n})={\frac {n!}{(1+x)^{n+1}}}$
HINT. Reduce fraction to form $-1+{\frac {2}{1+x}}$ before differentiating.
22. If $y=e^{x}\sin x$, prove that ${\frac {d^{2}y}{dx^{2}}}-2{\frac {dy}{dx}}+2y=0$.
23. If $y=a\cos(\log x)+b\sin(\log x)$, prove that $x^{2}{\frac {d^{2}y}{dx^{2}}}+x{\frac {dy}{dx}}+y=0$.

Use Leibnitz's Formula in the next four examples:
24. $y=x^{2}a^{x}$.	${\frac {d^{n}y}{dx^{n}}}=a^{x}(\log a)^{n-2}[(x\log a+n)^{2}-n]$
25. $y=e^{x}x$.	${\frac {d^{n}y}{dx^{n}}}=e^{x}(x+n)$.
26. $f(x)=e^{x}\sin x$.	$f^{(n)}(x)=({\sqrt {2}})^{n}e^{x}sin\left(x+{\frac {n\pi }{4}}\right)$.
27. $f(\theta )=\cos a\theta \cos b\theta$.	$f^{n}(\theta )={\frac {(a+b)^{n}}{2}}\cos \left[(a+b)\theta +{\frac {n\pi }{2}}\right]$ $+{\frac {(a-b)^{n}}{2}}\cos \left[(a-b)\theta +{\frac {n\pi }{2}}\right]$.
28. Show that the formulas for acceleration, (14), (15), §71, may be written
$\alpha ={\frac {d^{2}s}{dt^{2}}},\alpha _{x}={\frac {d^{2}x}{dt^{2}}},\alpha _{y}={\frac {d^{2}y}{dt^{2}}}$.
29. $y^{2}=4ax$.	${\frac {d^{2}y}{dx^{2}}}=-{\frac {4a^{2}}{y^{3}}}$.
30. $b^{2}x^{2}+a^{2}Y^{2}=a^{2}b^{2}$.	${\frac {d^{2}y}{dx^{2}}}=-{\frac {b^{4}}{a^{2}y^{3}}};{\frac {d^{3}y}{dx^{2}}}=-{\frac {3b^{6}x}{a^{4}y^{5}}}$.
31. $x^{2}+y^{2}=r^{2}$.	${\frac {d^{2}y}{dx^{2}}}=-{\frac {r^{2}}{y^{3}}}$.
32. $y^{2}+y=x^{2}$	${\frac {d^{3}y}{dx^{3}}}=-{\frac {24x}{(1+2y)^{5}}}$.
33. $ax^{2}+2hxy+by^{2}=1$.	${\frac {d^{2}y}{dx^{2}}}={\frac {h^{2}-ab}{(hx+by)^{3}}}$.
34. $y^{2}-2xy=a^{2}$.	${\frac {d^{2}y}{dx^{2}}}={\frac {a^{2}}{(y-x)^{3}}};{\frac {d^{3}y}{dx^{3}}}=-{\frac {3a^{2}x}{(y-x)^{5}}}$.
35. $\sec \phi \cos \theta =c$.	${\frac {d^{2}\theta }{d\phi ^{2}}}={\frac {\tan ^{2}\theta -\tan ^{2}\phi }{\tan ^{3}\theta }}$.
36. $\theta \tan(\phi +\theta )$.	${\frac {d^{3}\theta }{d\phi ^{3}}}=-{\frac {2(5+8\theta ^{2}+3\theta ^{4})}{\theta ^{8}}}$.
37. Find the second derivative in the following:
(a) $\log(u+v)=u-v$.	(e) $y^{3}+x^{3}-3axy=0$.
(b) $e^{u}+u=e^{v}+v$.	(f) $y^{2}-2mxy+x^{2}-a=0$.
(c) $s=1+te^{s}$.	(g) $y=\sin(x+y)$.
(d) $e^{s}+st-e=0$.	(h) $e^{x+y}=xy$.

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1. To make this correspondence complete, u and v are considered as ${\frac {d^{0}u}{dx^{0}}}$ and ${\frac {d^{0}v}{ddx^{0}}}$.

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