Page:Calculus Made Easy.pdf/61

Then

${\frac {dy}{dx}}={\frac {(x^{2}+a)\;{\frac {d(bx^{5}+c)}{dx}}-(bx^{5}+c)\;{\frac {d(x^{2}+a)}{dx}}}{(x^{2}+a)^{2}}}$

$={\frac {(x^{2}+a)(5bx^{4})-(bx^{5}+c)(2x)}{(x^{2}+a)^{2}}}$,

${\frac {dy}{dx}}={\frac {3bx^{6}+5abx^{4}-2cx}{(x^{2}+a)^{2}}}$. (Answer.)

The working out of quotients is often tedious, but there is nothing difficult about it.

Some further examples fully worked out are given hereafter.

(1) Differentiate $y={\dfrac {a}{b^{2}}}x^{3}-{\frac {a^{2}}{b}}x+{\frac {a^{2}}{b^{2}}}$.

Being a constant, $a^{2}b^{2}$ vanishes, and we have

${\frac {dy}{dx}}={\frac {a}{b^{2}}}\times 3\times x^{3-1}-{\frac {a^{2}}{b}}\times 1\times x^{1-1}$.

But $x^{1-1}=x^{0}=1$; so we get:

${\frac {dy}{dx}}={\frac {3a}{b^{2}}}x^{2}-{\frac {a^{2}}{b}}$.

(2) Differentiate $y=2a{\sqrt {bx^{3}}}-{\frac {3b{\sqrt[{3}]{a}}}{x}}-2{\sqrt {ab}}$.

Putting $x$ in the index form, we get

$y=2a{\sqrt {b}}x^{\frac {3}{2}}-3b{\sqrt[{3}]{a}}x^{-1}-2{\sqrt {ab}}.$.

Now

${\frac {dy}{dx}}=2a{\sqrt {b}}\times {\tfrac {3}{2}}\times x^{{\frac {3}{2}}-1}-3b{\sqrt[{3}]{a}}\times (-1)\times x^{-1-1}$;

or,

${\frac {dy}{dx}}=3a{\sqrt {bx}}+{\frac {3b{\sqrt[{3}]{a}}}{x^{2}}}$.