Calculus Made Easy/Chapter 9

by Silvanus Phillips Thompson

CHAPTER IX.

INTRODUCING A USEFUL DODGE.

Sometimes one is stumped by finding that the expression to be differentiated is too complicated to tackle directly.

Thus, the equation

$y=(x^{2}+a^{2})^{\tfrac {3}{2}}$

is awkward to a beginner.

Now the dodge to turn the difficulty is this: Write some symbol, such as $u$, for the expression $x^{2}+a^{2}$; then the equation becomes

$y=u^{\tfrac {3}{2}}$,

which you can easily manage; for

${\frac {dy}{du}}={\frac {3}{2}}u^{\tfrac {1}{2}}$.

Then tackle the expression

$u=x^{2}+a^{2}$,

and differentiate it with respect to $x$,

${\frac {du}{dx}}=2x$.

Then all that remains is plain sailing;

for	${\frac {dy}{dx}}$	$={\frac {dy}{du}}\times {\frac {du}{dx}}$;
that is,	${\frac {dy}{dx}}$	$={\frac {3}{2}}u^{\tfrac {1}{2}}\times {2x}$
		$={\tfrac {3}{2}}(x^{2}+a^{2})^{\tfrac {1}{2}}\times {2x}$
		$=3x(x^{2}+a^{2})^{\tfrac {1}{2}}$;

and so the trick is done.

By and by, when you have learned how to deal with sines, and cosines, and exponentials, you will find this dodge of increasing usefulness.

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Examples.

Let us practise this dodge on a few examples.

(1) Differentiate $y={\sqrt {a+x}}$.

Let $a+x=u$.

${\frac {du}{dx}}=1;\quad y=u^{\tfrac {1}{2}};\quad {\frac {dy}{du}}={\tfrac {1}{2}}u^{-{\tfrac {1}{2}}}={\tfrac {1}{2}}(a+x)^{-{\tfrac {1}{2}}}$.
${\frac {dy}{dx}}={\frac {dy}{du}}\times {\frac {du}{dx}}={\frac {1}{2{\sqrt {a+x}}}}$.

(2) Differentiate $y={\dfrac {1}{\sqrt {a+x^{2}}}}$.

Let $a+x^{2}=u$.

${\frac {du}{dx}}=2x;\quad y=u^{-{\frac {1}{2}}};\quad {\frac {dy}{du}}=-{\tfrac {1}{2}}u^{-{\frac {3}{2}}}$.
${\frac {dy}{dx}}={\frac {dy}{du}}\times {\frac {du}{dx}}=-{\frac {x}{\sqrt {(a+x^{2})^{3}}}}$.

(3) Differentiate $y=\left(m-nx^{\frac {2}{3}}+{\dfrac {p}{x^{\frac {4}{3}}}}\right)^{a}$.

Let $m-nx^{\frac {2}{3}}+px^{-{\frac {4}{3}}}=u$.

${\frac {du}{dx}}=-{\tfrac {2}{3}}nx^{-{\frac {1}{3}}}-{\tfrac {4}{3}}px^{-{\frac {7}{3}}}$;

$y=u^{a};\quad {\frac {dy}{du}}=au^{a-1}$.

${\frac {dy}{dx}}={\frac {dy}{du}}\times {\frac {du}{dx}}=-a\left(m-nx^{\frac {2}{3}}+{\frac {p}{x^{\frac {4}{3}}}}\right)^{a-1}({\tfrac {2}{3}}nx^{-{\frac {1}{3}}}+{\tfrac {4}{3}}px^{-{\frac {7}{3}}}).$

(4) Differentiate $y={\dfrac {1}{\sqrt {x^{3}-a^{2}}}}$.

Let $u=x^{3}-a^{2}$.

${\frac {du}{dx}}=3x^{2};\quad y=u^{-{\frac {1}{2}}};\quad {\frac {dy}{du}}=-{\frac {1}{2}}(x^{3}-a^{2})^{-{\frac {3}{2}}}$.
${\frac {dy}{dx}}={\frac {dy}{du}}\times {\frac {du}{dx}}=-{\frac {3x^{2}}{2{\sqrt {(x^{3}-a^{2})^{3}}}}}$.

(5) Differentiate $y={\sqrt {\dfrac {1-x}{1+x}}}$.

Write this as $y={\dfrac {(1-x)^{\frac {1}{2}}}{(1+x)^{\frac {1}{2}}}}$.

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

(We may also write $y=(1-x)^{\frac {1}{2}}(1+x)^{-{\frac {1}{2}}}$ and differentiate as a product.)

Proceeding as in example (1) above, we get

	${\frac {dy}{dx}}$	$=-{\frac {(1+x)^{\frac {1}{2}}}{2(1+x){\sqrt {1-x}}}}-{\frac {(1-x)^{\frac {1}{2}}}{2(1+x){\sqrt {1+x}}}}$
		$=-{\frac {1}{2{\sqrt {1+x}}{\sqrt {1-x}}}}-{\frac {\sqrt {1-x}}{2{\sqrt {(1+x)^{3}}}}}$;
or	${\frac {dy}{dx}}$	$=-{\frac {1}{(1+x){\sqrt {1-x^{2}}}}}$.

(6) Differentiate $y={\sqrt {\dfrac {x^{3}}{1+x^{2}}}}$.

We may write this

$y=x^{\tfrac {3}{2}}(1+x^{2})^{-{\tfrac {1}{2}}}$;

${\frac {dy}{dx}}={\tfrac {3}{2}}x^{\tfrac {1}{2}}(1+x^{2})^{-{\tfrac {1}{2}}}+x^{\tfrac {3}{2}}\times {\frac {d\left[(1+x^{2})^{-{\tfrac {1}{2}}}\right]}{dx}}$.

Differentiating $(1+x^{2})^{-{\frac {1}{2}}}$, as shown in example (2) above, we get

${\frac {d\left[(1+x^{2})^{-{\tfrac {1}{2}}}\right]}{dx}}=-{\frac {x}{\sqrt {(1+x^{2})^{3}}}}$;

so that

${\frac {dy}{dx}}={\frac {3{\sqrt {x}}}{2{\sqrt {1+x^{2}}}}}-{\frac {\sqrt {x^{5}}}{\sqrt {(1+x^{2})^{3}}}}={\frac {{\sqrt {x}}(3+x^{2})}{2{\sqrt {(1+x^{2})^{3}}}}}$.

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

Let $x+{\sqrt {x^{2}+x+a}}=u$.

${\frac {du}{dx}}=1+{\frac {d\left[(x^{2}+x+a)^{\tfrac {1}{2}}\right]}{dx}}$.

$y=u^{3}$; and ${\frac {dy}{du}}=3u^{2}=3\left(x+{\sqrt {x^{2}+x+a}}\right)^{2}$.

Now let $(x^{2}+x+a)^{\tfrac {1}{2}}=v$ and $(x^{2}+x+a)=w$.

	${\frac {dw}{dx}}$	$=2x+1;\quad v=w^{\frac {1}{2}};\quad {\frac {dv}{dw}}={\tfrac {1}{2}}w^{-{\frac {1}{2}}}$.
	${\frac {dv}{dx}}$	$={\frac {dv}{dw}}\times {\frac {dw}{dx}}={\tfrac {1}{2}}(x^{2}+x+a)^{-{\tfrac {1}{2}}}(2x+1)$.
Hence	${\frac {du}{dx}}$	$=1+{\frac {2x+1}{2{\sqrt {x^{2}+x+a}}}}$,
	${\frac {dy}{dx}}$	$={\frac {dy}{du}}\times {\frac {du}{dx}}$
		$=3\left(x+{\sqrt {x^{2}+x+a}}\right)^{2}\left(1+{\frac {2x+1}{2{\sqrt {x^{2}+x+a}}}}\right).$

(8) Differentiate $y={\sqrt {\dfrac {a^{2}+x^{2}}{a^{2}-x^{2}}}}{\sqrt[{3}]{\dfrac {a^{2}-x^{2}}{a^{2}+x^{2}}}}$.

We get

$y$	$={\frac {(a^{2}+x^{2})^{\frac {1}{2}}(a^{2}-x^{2})^{\frac {1}{3}}}{(a^{2}-x^{2})^{\frac {1}{2}}(a^{2}+x^{2})^{\frac {1}{3}}}}=(a^{2}+x^{2})^{\frac {1}{6}}(a^{2}-x^{2})^{-{\frac {1}{6}}}$.
${\frac {dy}{dx}}$	$=(a^{2}+x^{2})^{\tfrac {1}{6}}{\frac {d[(a^{2}-x^{2})^{-{\tfrac {1}{6}}}]}{dx}}+{\frac {d[(a^{2}+x^{2})^{\tfrac {1}{6}}]}{(a^{2}-x^{2})^{\tfrac {1}{6}}\,dx}}$.

Let $u=(a^{2}-x^{2})^{-{\frac {1}{6}}}$ and $v=(a^{2}-x^{2})$.

$u$	$=v^{-{\frac {1}{6}}};\quad {\frac {du}{dv}}=-{\frac {1}{6}}v^{-{\frac {7}{6}}};\quad {\frac {dv}{dx}}=-2x$.
${\frac {du}{dx}}$	$={\frac {du}{dv}}\times {\frac {dv}{dx}}={\frac {1}{3}}x(a^{2}-x^{2})^{-{\frac {7}{6}}}$.

Let $w=(a^{2}-x^{2})^{\frac {1}{6}}$ and $z=(a^{2}+x^{2})$.

$w$	$=z^{\frac {1}{6}};\quad {\frac {dw}{dz}}={\frac {1}{6}}z^{-{\frac {5}{6}}};\quad {\frac {dz}{dx}}=2x$.
${\frac {dw}{dx}}$	$={\frac {dw}{dz}}\times {\frac {dz}{dx}}={\frac {1}{3}}x(a^{2}+x^{2})^{-{\frac {5}{6}}}$.

Hence

	${\frac {dy}{dx}}$	$=(a^{2}+x^{2})^{\frac {1}{6}}{\frac {x}{3(a^{2}-x^{2})^{\frac {7}{6}}}}+{\frac {x}{3(a^{2}-x^{2})^{\frac {1}{6}}(a^{2}+x^{2})^{\frac {5}{6}}}}$;
or	${\frac {dy}{dx}}$	$={\frac {x}{3}}\left[{\sqrt[{6}]{\frac {a^{2}+x^{2}}{(a^{2}-x^{2})^{7}}}}+{\frac {1}{\sqrt[{6}]{(a^{2}-x^{2})(a^{2}+x^{2})^{5}]}}}\right]$.

(9) Differentiate $y^{n}$ with respect to $y^{5}$.

${\frac {d(y^{n})}{d(y^{5})}}={\frac {ny^{n-1}}{5y^{5-1}}}={\frac {n}{5}}y^{n-5}$.

(10) Find the first and second differential coefficients of $y={\dfrac {x}{b}}{\sqrt {(a-x)x}}$.

${\frac {dy}{dx}}={\frac {x}{b}}\,{\frac {d{\bigl \{}{\bigl [}(a-x)x{\bigr ]}^{\frac {1}{2}}{\bigr \}}}{dx}}+{\frac {\sqrt {(a-x)x}}{b}}$.

Let ${\bigl [}(a-x)x{\bigr ]}^{\frac {1}{2}}=u$ and let $(a-x)x=w$; then $u=w^{\frac {1}{2}}$.

${\frac {du}{dw}}={\frac {1}{2}}w^{-{\frac {1}{2}}}={\frac {1}{2w^{\frac {1}{2}}}}={\frac {1}{2{\sqrt {(a-x)x}}}}$.

${\frac {dw}{dx}}=a-2x$.

${\frac {du}{dw}}\times {\frac {dw}{dx}}={\frac {du}{dx}}={\frac {a-2x}{2{\sqrt {(a-x)x}}}}$.

Hence

${\frac {dy}{dx}}={\frac {x(a-2x)}{2b{\sqrt {(a-x)x}}}}+{\frac {\sqrt {(a-x)x}}{b}}={\frac {x(3a-4x)}{2b{\sqrt {(a-x)x}}}}$.

Now

${\frac {d^{2}y}{dx^{2}}}$	$={\frac {2b{\sqrt {(a-x)x}}\,(3a-8x)-{\dfrac {(3ax-4x^{2})b(a-2x)}{\sqrt {(a-x)x}}}}{4b^{2}(a-x)x}}$
${\displaystyle }$	$={\frac {du}{dv}}\times {\frac {dv}{dx}}={\frac {1}{3}}x(a^{2}-x^{2})^{-{\frac {7}{6}}}$.

(We shall need these two last differential coefficients later on. See Ex. X. No. 11)

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Exercises VI. (See page 257 for Answers.)

Differentiate the following:

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

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

(3) $y={\dfrac {1}{\sqrt {a+x}}}$.

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

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

(6) $y={\dfrac {\sqrt[{3}]{x^{4}+a}}{\sqrt[{2}]{x^{3}+a}}}$.

(7) $y={\dfrac {a^{2}+x^{2}}{(a+x)^{2}}}$.

(8) Differentiate $y^{5}$ with respect to $y^{2}$.

(9) Differentiate $y={\dfrac {\sqrt {1-\theta ^{2}}}{1-\theta }}$.

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The process can be extended to three or more differential coefficients, so that ${\dfrac {dy}{dx}}={\dfrac {dy}{dz}}\times {\dfrac {dz}{dv}}\times {\dfrac {dv}{dx}}$

Examples.

(1) If $z=3x^{4}$; $v={\dfrac {7}{z^{2}}}$; $y={\sqrt {1+v}}$, find ${\dfrac {dv}{dx}}$.

We have

${\frac {dy}{dv}}$	$={\frac {1}{2{\sqrt {1+v}}}};\quad {\frac {dv}{dz}}=-{\frac {14}{z^{3}}};\quad {\frac {dz}{dx}}=12x^{3}$.
${\frac {dy}{dx}}$	$=-{\frac {168x^{3}}{(2{\sqrt {1+v}})z^{3}}}=-{\frac {28}{3x^{5}{\sqrt {9x^{8}+7}}}}$.

(2) If $t={\dfrac {1}{5{\sqrt {\theta }}}}$; $x=t^{3}+{\dfrac {t}{2}}$; $v={\dfrac {7x^{2}}{\sqrt[{3}]{x-1}}}$, find ${\dfrac {d\phi }{dx}}$.

${\frac {dv}{dx}}={\frac {7x(5x-6)}{3{\sqrt[{3}]{(x-1)^{4}}}}};\quad {\frac {dx}{dt}}=3t^{2}+{\frac {1}{2}};\quad {\frac {dt}{d\theta }}=-{\frac {1}{10{\sqrt {\theta ^{3}}}}}$.

Hence ${\frac {dv}{d\theta }}=-{\frac {7x(5x-6)(3t^{2}+{\frac {1}{2}})}{30{\sqrt[{3}]{(x-1)^{4}}}{\sqrt {\theta ^{3}}}}}$,

an expression in which $x$ must be replaced by its value, and $t$ by its value in terms of $\theta$.

(3) If $\theta ={\dfrac {3a^{2}x}{\sqrt {x^{3}}}}$; $\omega ={\dfrac {\sqrt {1-\theta ^{2}}}{1+\theta }}$; and $\phi ={\sqrt {3}}-{\dfrac {1}{\omega {\sqrt {2}}}}$, find ${\dfrac {d\phi }{dx}}$.

We get

$\theta =3a^{2}x^{-{\frac {1}{2}}};\quad \omega ={\sqrt {\frac {1-\theta }{1+\theta }}};\quad$ and $\quad \phi ={\sqrt {3}}-{\frac {1}{\sqrt {2}}}\omega ^{-1}$.

${\frac {d\theta }{dx}}=-{\frac {3a^{2}}{2{\sqrt {x^{3}}}}};\quad {\frac {d\omega }{d\theta }}=-{\frac {1}{(1+\theta ){\sqrt {1-\theta ^{2}}}}}$

(see example 5, p. 69); and

${\frac {d\phi }{d\omega }}={\frac {1}{{\sqrt {2}}\omega ^{2}}}$.

So that ${\dfrac {d\theta }{dx}}={\dfrac {1}{{\sqrt {2}}\times \omega ^{2}}}\times {\dfrac {1}{(1+\theta ){\sqrt {1-\theta ^{2}}}}}\times {\dfrac {3a^{2}}{2{\sqrt {x^{3}}}}}$ Replace now first $\omega$, then $\theta$ by its value.

 

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Exercises VII

You can now successfully try the following. (See page 257 for Answers.)

(1) If $u={\frac {1}{2}}x^{3}$; $v=3(u+u^{2})$; and $w={\dfrac {1}{v^{2}}}$, find ${\dfrac {dw}{dx}}$.

(2) If $y=3x^{2}+{\sqrt {2}}$; $z={\sqrt {1+y}}$; and $v={\dfrac {1}{{\sqrt {3}}+4z}}$, find ${\dfrac {dv}{dx}}$.

(3) If $y={\dfrac {x^{3}}{\sqrt {3}}}$; $z=(1+y)^{2}$; and $u={\dfrac {1}{\sqrt {1+z}}}$, find ${\dfrac {du}{dx}}$.

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