Page:Calculus Made Easy.pdf/93

${\frac {dw}{dx}}=a-2x$.

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

Hence

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

Now

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

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

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Exercises VI. (See page 257 for Answers.)

Differentiate the following:

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

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

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

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

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

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

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