Question

the question says: "The rate of change of the function f(x)=secx+cosx is given by the expression secxtanx-sinx. Show that this expression can also be written as sinxtan(x)^2

Answer 1

Question 5606

The rate of change of the function $f(x) = \sec x + \cos x$ is given by the expression $\sec \tan x - \sin x$. Show that this expression can also be written as $\sin \tan(x)^2$

First, let's find derivative of $f(x)$ and then show easy operations to solve the given task. Remember, that

$$\sec (x) = \frac {1}{\cos (x)}; \tan (x) = \frac {\sin (x)}{\cos (x)}$$

The derivative

$$f ^ {\prime} (x) = - \sin (x) + \frac {1}{\cos^ {2} (x)} \cdot \sin (x)$$

According to (1)

$$\sec (x) \cdot \tan (x) = \frac {\sin (x)}{\cos^ {2} (x)}$$

Then, combining (2) and (3), we obtain: $f'(x) = \sec(x)\tan(x) - \sin(x)$ (4)

Also, we can convert (2), using $\sin^2(x) + \cos^2(x) = 1$ :

$$f ^ {\prime} (x) = \frac {- \sin (x) \cos^ {2} (x) + \sin (x)}{\cos^ {2} (x)} = \frac {\sin (x) [ 1 - \cos^ {2} (x) ]}{\cos^ {2} (x)} = \frac {\sin^ {3} (x)}{\cos^ {2} (x)} = \sin (x) \cdot \tan^ {2} (x)$$

We have proved both equalities.