Page:Calculus Made Easy.pdf/89

 (3) Differentiate $$y=\left(m-nx^{\frac{2}{3}}+\dfrac{p}{x^{\frac{4}{3}}}\right)^a$$.

Let $$m-nx^{\frac{2}{3}}+px^{-\frac{4}{3}}=u$$.

$$\frac{du}{dx} = -\tfrac{2}{3} nx^{-\frac{1}{3}} - \tfrac{4}{3} px^{-\frac{7}{3}}$$;

$$y = u^a;\quad \frac{dy}{du} = a u^{a-1}$$.

$$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = -a\left(m -nx^{\frac{2}{3}} + \frac{p}{x^{\frac{4}{3}}}\right)^{a-1} (\tfrac{2}{3} nx^{-\frac{1}{3}} + \tfrac{4}{3} px^{-\frac{7}{3}}).$$

(4) Differentiate $$y=\dfrac{1}{\sqrt{x^3 - a^2}}$$.

Let $$u=x^3-a^2$$.

$$ \frac{du}{dx} = 3x^2;\quad y = u^{-\frac{1}{2}};\quad \frac{dy}{du}=-\frac{1}{2}(x^3 - a^2)^{-\frac{3}{2}}$$. $$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = -\frac{3x^2}{2\sqrt{(x^3 - a^2)^3}} $$.

(5) Differentiate $$y=\sqrt{\dfrac{1-x}{1+x}}$$.

Write this as $$y=\dfrac{(1-x)^{\frac{1}{2}}}{(1+x)^{\frac{1}{2}}}$$.

(We may also write $$y=(1-x)^{\frac{1}{2}}(1+x)^{-\frac{1}{2}}$$ and differentiate as a product.)