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} $|undefined $=\frac{(x^2 + a)(5bx^4) - (bx^5 + c)(2x)}{(x^2 + a)^2}$, $\frac{dy}{dx}=\frac{3bx^6 + 5abx^4 - 2cx}{(x^2 + a)^2}$. (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, $$a^2b^2$$ 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 $y = 2a\sqrt{b} x^{\frac{3}{2}} - 3b \sqrt[3]{a} x^{-1} - 2\sqrt{ab}.$.|undefined

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