Page:Calculus Made Easy.pdf/95

We get

$\theta =3a^{2}x^{-{\frac {1}{2}}};\quad \omega ={\sqrt {\frac {1-\theta }{1+\theta }}};\quad$ and $\quad \phi ={\sqrt {3}}-{\frac {1}{\sqrt {2}}}\omega ^{-1}$.

${\frac {d\theta }{dx}}=-{\frac {3a^{2}}{2{\sqrt {x^{3}}}}};\quad {\frac {d\omega }{d\theta }}=-{\frac {1}{(1+\theta ){\sqrt {1-\theta ^{2}}}}}$

(see example 5, p. 69); and

${\frac {d\phi }{d\omega }}={\frac {1}{{\sqrt {2}}\omega ^{2}}}$.

So that ${\dfrac {d\theta }{dx}}={\dfrac {1}{{\sqrt {2}}\times \omega ^{2}}}\times {\dfrac {1}{(1+\theta ){\sqrt {1-\theta ^{2}}}}}\times {\dfrac {3a^{2}}{2{\sqrt {x^{3}}}}}$ Replace now first $\omega$, then $\theta$ by its value.

 

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Exercises VII

You can now successfully try the following. (See page 257 for Answers.)

(1) If $u={\frac {1}{2}}x^{3}$; $v=3(u+u^{2})$; and $w={\dfrac {1}{v^{2}}}$, find ${\dfrac {dw}{dx}}$.

(2) If $y=3x^{2}+{\sqrt {2}}$; $z={\sqrt {1+y}}$; and $v={\dfrac {1}{{\sqrt {3}}+4z}}$, find ${\dfrac {dv}{dx}}$.

(3) If $y={\dfrac {x^{3}}{\sqrt {3}}}$; $z=(1+y)^{2}$; and $u={\dfrac {1}{\sqrt {1+z}}}$, find ${\dfrac {du}{dx}}$.