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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

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

is awkward to a beginner.

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

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

which you can easily manage; for

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

Then tackle the expression

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

and differentiate it with respect to ${\displaystyle x}$,

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

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

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.

Examples.

Let us practise this dodge on a few examples.

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

Let ${\displaystyle a+x=u}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Proceeding as in example (1) above, we get

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

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

We may write this

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

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

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

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

so that

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

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

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

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

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

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

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

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

We get

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

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

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

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

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

Hence

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

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

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

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

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

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

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

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

Hence

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

Now

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

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

Exercises VI. (See page 257 for Answers.)

Differentiate the following:

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

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

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

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

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

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

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

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

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

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

Examples.

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

We have

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

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

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

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

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

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

We get

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

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

(see example 5, p. 69); and

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

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

Exercises VII

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

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

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

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