Page:Calculus Made Easy.pdf/64

 (9) Differentiate $$y=\frac{x^2}{x^2+1}$$.

$\frac{dy}{dx}=\frac{(x^2+1)\,2x-x^2\times 2x}{(x^2+1)^2}=\frac{2x}{(x^2+1)^2}$.

(10) Differentiate $y=\frac{a+\sqrt{x}}{a-\sqrt{x}}$.|undefined

In the indexed form, $y=\frac{a+x^{\frac{1}{2}}}{a - x^{\frac{1}{2}}}$.|undefined

$\frac{dy}{dx}=\frac{(a-x^{\frac{1}{2}})(\tfrac{1}{2}x^{-\frac{1}{2}})-(a+x^{\frac{1}{2}})(-\tfrac{1}{2}x^{-\frac{1}{2}})}{(a-x^{\frac{1}{2}})^2}=\frac{a-x^{\frac{1}{2}}+a+x^{\frac{1}{2}}}{2(a - x^{\frac{1}{2}})^2 x^{\frac{1}{2}}}$;|undefined hence $\frac{dy}{dx}=\frac{a}{(a-\sqrt{x})^2\,\sqrt{x}}$.|undefined

(11) Differentiate $\theta=\frac{1-a\sqrt[3]{t^2}}{1+a \sqrt[2]{t^3}}$.|undefined

Now $\theta=\frac{1-at^{\frac{2}{3}}}{1+at^{\frac{3}{2}}}$.|undefined

(12) A reservoir of square cross-section has sides sloping at an angle of $$45^\circ$$ with the vertical. The side