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Calculus Made Easy

(9) Differentiate $y={\frac {x^{2}}{x^{2}+1}}$ .

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

(10) Differentiate

$y={\frac {a+{\sqrt {x}}}{a-{\sqrt {x}}}}$ .

In the indexed form,

$y={\frac {a+x^{\frac {1}{2}}}{a-x^{\frac {1}{2}}}}$ .

${\frac {dy}{dx}}={\frac {(a-x^{\frac {1}{2}})({\tfrac {1}{2}}x^{-{\frac {1}{2}}})-(a+x^{\frac {1}{2}})(-{\tfrac {1}{2}}x^{-{\frac {1}{2}}})}{(a-x^{\frac {1}{2}})^{2}}}={\frac {a-x^{\frac {1}{2}}+a+x^{\frac {1}{2}}}{2(a-x^{\frac {1}{2}})^{2}x^{\frac {1}{2}}}}$ ;

hence

${\frac {dy}{dx}}={\frac {a}{(a-{\sqrt {x}})^{2}\,{\sqrt {x}}}}$ .

(11) Differentiate

$\theta ={\frac {1-a{\sqrt[{3}]{t^{2}}}}{1+a{\sqrt[{2}]{t^{3}}}}}$ .

Now

$\theta ={\frac {1-at^{\frac {2}{3}}}{1+at^{\frac {3}{2}}}}$ .

${\frac {d\theta }{dt}}$	$={\frac {(1+at^{\frac {3}{2}})(-{\tfrac {2}{3}}at^{-{\frac {1}{3}}})-(1-at^{\frac {2}{3}})\times {\tfrac {3}{2}}at^{\frac {1}{2}}}{(1+at^{\frac {3}{2}})^{2}}}$
	$={\frac {5a^{2}{\sqrt[{6}]{t^{7}}}-{\dfrac {4a}{\sqrt[{3}]{t}}}-9a{\sqrt[{2}]{t}}}{6(1+a{\sqrt[{2}]{t^{3}}})^{2}}}$ .

(12) A reservoir of square cross-section has sides sloping at an angle of $45^{\circ }$ with the vertical. The side

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