Page:Calculus Made Easy.pdf/94

(8) Differentiate $y^{5}$ with respect to $y^{2}$.

(9) Differentiate $y={\dfrac {\sqrt {1-\theta ^{2}}}{1-\theta }}$.

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The process can be extended to three or more differential coefficients, so that ${\dfrac {dy}{dx}}={\dfrac {dy}{dz}}\times {\dfrac {dz}{dv}}\times {\dfrac {dv}{dx}}$

Examples.

(1) If $z=3x^{4}$; $v={\dfrac {7}{z^{2}}}$; $y={\sqrt {1+v}}$, find ${\dfrac {dv}{dx}}$.

We have

${\frac {dy}{dv}}$	$={\frac {1}{2{\sqrt {1+v}}}};\quad {\frac {dv}{dz}}=-{\frac {14}{z^{3}}};\quad {\frac {dz}{dx}}=12x^{3}$.
${\frac {dy}{dx}}$	$=-{\frac {168x^{3}}{(2{\sqrt {1+v}})z^{3}}}=-{\frac {28}{3x^{5}{\sqrt {9x^{8}+7}}}}$.

(2) If $t={\dfrac {1}{5{\sqrt {\theta }}}}$; $x=t^{3}+{\dfrac {t}{2}}$; $v={\dfrac {7x^{2}}{\sqrt[{3}]{x-1}}}$, find ${\dfrac {d\phi }{dx}}$.

${\frac {dv}{dx}}={\frac {7x(5x-6)}{3{\sqrt[{3}]{(x-1)^{4}}}}};\quad {\frac {dx}{dt}}=3t^{2}+{\frac {1}{2}};\quad {\frac {dt}{d\theta }}=-{\frac {1}{10{\sqrt {\theta ^{3}}}}}$.

Hence ${\frac {dv}{d\theta }}=-{\frac {7x(5x-6)(3t^{2}+{\frac {1}{2}})}{30{\sqrt[{3}]{(x-1)^{4}}}{\sqrt {\theta ^{3}}}}}$,

an expression in which $x$ must be replaced by its value, and $t$ by its value in terms of $\theta$.

(3) If $\theta ={\dfrac {3a^{2}x}{\sqrt {x^{3}}}}$; $\omega ={\dfrac {\sqrt {1-\theta ^{2}}}{1+\theta }}$; and $\phi ={\sqrt {3}}-{\dfrac {1}{\omega {\sqrt {2}}}}$, find ${\dfrac {d\phi }{dx}}$.