Page:Calculus Made Easy.pdf/95

This page has been proofread, but needs to be validated.

INTRODUCING A USEFUL DODGE

75

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.

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 ${\dfrac {du}{dx}}$ .

This article is issued from Wikisource. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.