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Calculus Made Easy

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

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

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