How do find the derivative of #y= (1- sec x)/ tan x#?

Question

Charles Anctil · Accepted Answer

#(dy)/(dx)=cscx(cotx-cscx)#

Here ,

#y=(1-secx)/tanx#

#=>y=1/tanx-secx/tanx#

#=>y=cotx-(1/cosx)/(sinx/cosx)#

#=>y=cotx-cscx#

#(dy)/(dx)=-csc^2x-(-cscxcotx)#

#(dy)/(dx)=-csc^2x+cscxcotx#

#:.(dy)/(dx)=cscx(cotx-cscx)#

Paisley Johnson · Answer

undefined To find the derivative of $y = \frac{1 - \sec(x)}{\tan(x)}$, we use the quotient rule. The quotient rule states that if $u$ and $v$ are differentiable functions of $x$, then the derivative of $\frac{u}{v}$ is $\frac{u'v - uv'}{v^2}$. Applying the quotient rule to the given function, we have:

$$y' = \frac{(\tan(x) \cdot \sec(x) - (1 - \sec(x)) \cdot \sec^2(x))}{\tan^2(x)}$$

Simplify the expression:

$$y' = \frac{\tan(x) \cdot \sec(x) - (1 - \sec(x)) \cdot \sec^2(x)}{\tan^2(x)}$$

HIX Tutor · Answer

undefined To find the derivative of y=(1-sec(x))/tan(x), we can use the quotient rule.

Step 1: Recall that the quotient rule states that if we have a function f(x) divided by a function g(x), the derivative is given by (g(x) * f'(x) - f(x) * g'(x))/[g(x)]^2.

Step 2: Let's start by finding f(x) and g(x). In this case, f(x) = 1 - sec(x) and g(x) = tan(x).

Step 3: Now, we need to find f'(x) and g'(x), which are the derivatives of f(x) and g(x).

- The derivative of f(x) = 1 - sec(x) with respect to x is f'(x) = 0 - sec(x) * tan(x).
- The derivative of g(x) = tan(x) with respect to x is g'(x) = sec^2(x).

Step 4: Now, substitute these values into the quotient rule formula:
(g(x) * f'(x) - f(x) * g'(x))/[g(x)]^2 = [tan(x) * (0 - sec(x) * tan(x)) - (1 - sec(x)) * sec^2(x)] / [tan(x)]^2

Step 5: Simplify:
= [(0 - sec^2(x) * tan^2(x)) - (1 - sec(x)) * sec^2(x)] / tan^2(x)
= [-sec^2(x) * tan^2(x) - sec^2(x) + sec^3(x)] / tan^2(x)

So, the derivative of y=(1-sec(x))/tan(x) is:
[-sec^2(x) * tan^2(x) - sec^2(x) + sec^3(x)] / tan^2(x)