Page:Calculus Made Easy.pdf/91

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

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

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

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

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

We get

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