How do you find the derivative of $\frac{sec x + csc x}{csc x}$ $(secx+cscx)/cscx$ ?

Question

How do you find the derivative of $\frac{sec x + csc x}{csc x}$ $(secx+cscx)/cscx$ ?

2 Answers

Impact of this question

3988 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-06T10:14:37

$\frac{d}{d x} \frac{sec x + csc x}{csc x} = {sec}^{2} x$ $d/(dx)(secx+cscx)/cscx=sec^2x$

Explanation:

Let us first simplify $\frac{sec x + csc x}{csc x}$ $(secx+cscx)/cscx$

= $\frac{\frac{1}{cos x} + \frac{1}{sin}}{\frac{1}{sin x}}$ $(1/cosx+1/sin)/(1/sinx)$

= $(\frac{1}{cos x} + \frac{1}{sin}) \times sin x$ $(1/cosx+1/sin)xxsinx$

= $\frac{sin x}{cos x} + 1$ $sinx/cosx+1$

= $tan x + 1$ $tanx+1$

Hence, $\frac{d}{d x} \frac{sec x + csc x}{csc x}$ $d/(dx)(secx+cscx)/cscx$

= $\frac{d}{d x} (tan x + 1)$ $d/(dx)(tanx+1)$

= ${sec}^{2} x + 0$ $sec^2x+0$

= ${sec}^{2} x$ $sec^2x$

Answer 2 · 2017-03-06T10:18:01

Since, $\frac{sec x + csc x}{csc x} = \frac{sec x}{csc x} + \frac{csc x}{csc x} = \frac{\frac{1}{cos x}}{\frac{1}{sin x}} + 1$ $(secx+cscx)/cscx=secx/cscx+cscx/cscx=(1/cosx)/(1/sinx) +1$

$= \frac{sin x}{cos x} + 1 = tan x + 1,$ $=sinx/cosx+1=tanx+1,$

$\frac{d}{d x} {\frac{sec x + csc x}{csc x}} = \frac{d}{d x} (tan x + 1) = {sec}^{2} x .$ $d/dx{(secx+cscx)/cscx}=d/dx(tanx+1)=sec^2x.$

Science

Math

Humanities

... and beyond

How do you find the derivative of $\frac{sec x + csc x}{csc x}$ $(secx+cscx)/cscx$ ?

2 Answers

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the derivative of secx+cscxcscx(secx+cscx)/cscx?

2 Answers

Explanation:

Related questions

Impact of this question

How do you find the derivative of $\frac{sec x + csc x}{csc x}$ $(secx+cscx)/cscx$ ?