Page:Calculus Made Easy.pdf/172

(7) $y=(x+3)^{2}{\sqrt {x-2}}$.

${\begin{aligned}\log _{\epsilon }y&=2\log _{\epsilon }(x+3)+{\tfrac {1}{2}}\log _{\epsilon }(x-2).\\{\frac {1}{y}}\,{\frac {dy}{dx}}&={\frac {2}{(x+3)}}+{\frac {1}{2(x-2)}};\\{\frac {dy}{dx}}&=(x+3)^{2}{\sqrt {x-2}}\left\{{\frac {2}{x+3}}+{\frac {1}{2(x-2)}}\right\}.\end{aligned}}$

(8) $y=(x^{2}+3)^{3}(x^{3}-2)^{\frac {2}{3}}$.

${\begin{aligned}\log _{\epsilon }y&=3\log _{\epsilon }(x^{2}+3)+{\tfrac {2}{3}}\log _{\epsilon }(x^{3}-2);\\{\frac {1}{y}}\,{\frac {dy}{dx}}&=3{\frac {2x}{(x^{2}+3)}}+{\frac {2}{3}}{\frac {3x^{2}}{x^{3}-2}}={\frac {6x}{x^{2}+3}}+{\frac {2x^{2}}{x^{3}-2}}.\end{aligned}}$

(For if $y=\log _{\epsilon }(x^{2}+3)$, let $x^{2}+3=z$ and $u=\log _{\epsilon }z$.

${\frac {du}{dz}}={\frac {1}{z}};\quad {\frac {dz}{dx}}=2x;\quad {\frac {du}{dx}}={\frac {2x}{x^{2}+3}}.$

Similarly, if $v=\log _{\epsilon }(x^{3}-2)$, ${\dfrac {dv}{dx}}={\dfrac {3x^{2}}{x^{3}-2}}$) and

${\frac {dy}{dx}}=(x^{2}+3)^{3}(x^{3}-2)^{\frac {2}{3}}\left\{{\frac {6x}{x^{2}+3}}+{\frac {2x^{2}}{x^{3}-2}}\right\}.$

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

${\begin{aligned}\log _{\epsilon }y&={\frac {1}{2}}\log _{\epsilon }(x^{2}+a)-{\frac {1}{3}}\log _{\epsilon }(x^{3}-a).\\{\frac {1}{y}}\,{\frac {dy}{dx}}&={\frac {1}{2}}\,{\frac {2x}{x^{2}+a}}-{\frac {1}{3}}\,{\frac {3x^{2}}{x^{3}-a}}={\frac {x}{x^{2}+a}}-{\frac {x^{2}}{x^{3}-a}}\\and\quad \quad \quad {\frac {dy}{dx}}&={\frac {\sqrt[{2}]{x^{2}+a}}{\sqrt[{3}]{x^{3}-a}}}\left\{{\frac {x}{x^{2}+a}}-{\frac {x^{2}}{x^{3}-a}}\right\}.\end{aligned}}$