Page:Calculus Made Easy.pdf/193

 (7) $$y = \sqrt{1+3\tan^2\theta}$$; $$y=(1+3 \tan^2 \theta)^{\frac{1}{2}}$$.

Let $$3\tan^2\theta=v$$.

(for, if $$\tan \theta = u$$,

(8) $$y=\sin x \cos x$$.

Exercises XIV. (See page 261 for Answers.)

(1) Differentiate the following:

(2) Find the value of $$\theta$$ for which $$\sin\theta \times \cos\theta$$ is a maximum.

(3) Differentiate $$y=\dfrac{1}{2\pi} \cos 2\pi nt$$.