Page:Calculus Made Easy.pdf/95

 We get

(see example 5, p. 69); and

So that $$\dfrac{d\theta}{dx} = \dfrac{1}{\sqrt{2} \times \omega^2} \times \dfrac{1}{(1+\theta) \sqrt{1-\theta^2}} \times \dfrac{3a^2}{2\sqrt{x^3}}$$ Replace now first $$\omega$$, then $$\theta$$ by its value.

Exercises VII

You can now successfully try the following. (See page 257 for Answers.)

(1) If $$u=\frac{1}{2}x^3$$; $$v=3(u+u^2)$$;  and $$w=\dfrac{1}{v^2}$$, find $$\dfrac{dw}{dx}$$.

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

(3) If $$y=\dfrac{x^3}{\sqrt{3}}$$; $$z=(1+y)^2$$; and $$u=\dfrac{1}{\sqrt{1+z}}$$, find $$\dfrac{du}{dx}$$.