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Calculus Made Easy

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.

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

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

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