Page:Calculus Made Easy.pdf/88

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

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