Page:Calculus Made Easy.pdf/91

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

Let $x+{\sqrt {x^{2}+x+a}}=u$.

${\frac {du}{dx}}=1+{\frac {d\left[(x^{2}+x+a)^{\tfrac {1}{2}}\right]}{dx}}$.

$y=u^{3}$; and ${\frac {dy}{du}}=3u^{2}=3\left(x+{\sqrt {x^{2}+x+a}}\right)^{2}$.

Now let $(x^{2}+x+a)^{\tfrac {1}{2}}=v$ and $(x^{2}+x+a)=w$.

	${\frac {dw}{dx}}$	$=2x+1;\quad v=w^{\frac {1}{2}};\quad {\frac {dv}{dw}}={\tfrac {1}{2}}w^{-{\frac {1}{2}}}$.
	${\frac {dv}{dx}}$	$={\frac {dv}{dw}}\times {\frac {dw}{dx}}={\tfrac {1}{2}}(x^{2}+x+a)^{-{\tfrac {1}{2}}}(2x+1)$.
Hence	${\frac {du}{dx}}$	$=1+{\frac {2x+1}{2{\sqrt {x^{2}+x+a}}}}$,
	${\frac {dy}{dx}}$	$={\frac {dy}{du}}\times {\frac {du}{dx}}$
		$=3\left(x+{\sqrt {x^{2}+x+a}}\right)^{2}\left(1+{\frac {2x+1}{2{\sqrt {x^{2}+x+a}}}}\right).$

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

We get

$y$	$={\frac {(a^{2}+x^{2})^{\frac {1}{2}}(a^{2}-x^{2})^{\frac {1}{3}}}{(a^{2}-x^{2})^{\frac {1}{2}}(a^{2}+x^{2})^{\frac {1}{3}}}}=(a^{2}+x^{2})^{\frac {1}{6}}(a^{2}-x^{2})^{-{\frac {1}{6}}}$.
${\frac {dy}{dx}}$	$=(a^{2}+x^{2})^{\tfrac {1}{6}}{\frac {d[(a^{2}-x^{2})^{-{\tfrac {1}{6}}}]}{dx}}+{\frac {d[(a^{2}+x^{2})^{\tfrac {1}{6}}]}{(a^{2}-x^{2})^{\tfrac {1}{6}}\,dx}}$.