How do you find the derivative of #y = secx/(1 + tanx)#?

Question

Isabella Ackerman · Accepted Answer

#dy/dx = (sinx - cosx)/(sinx + cosx)^2#

First of all, I'm assuming the function is #y = secx/(1 + tanx)#? Call your function #f(x)#.

#f(x) = secx/(1 + tanx)#

#f(x) = (1/cosx)/(1 + sinx/cosx)#

#f(x) = (1/cosx)/((cosx + sinx)/cosx)#

#f(x) = 1/cosx xx cosx/(cosx + sinx)#

#f(x) = 1/(cosx+ sinx)#

#f(x) = (cosx + sinx)^-1#

We let #y = u^-1# and #u = cosx + sinx#. Then #dy/(du) = -1u^(-2)# and #(du)/dx = -sinx + cosx#.

The chain rule states that #color(red)(dy/dx = dy/(du) xx (du)/dx#.

#dy/dx = -1/u^2 xx -sinx + cosx#

#dy/dx = (sinx - cosx)/(sinx + cosx)^2#

Hopefully this helps!

Samantha Adkison · Answer

undefined To find the derivative of $ y = \frac{\sec x}{1 + \tan x} $, we can use the quotient rule.

The quotient rule states that for functions $ u(x) $ and $ v(x) $ where $ v(x) \neq 0 $, the derivative of $ \frac{u(x)}{v(x)} $ is given by:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Using the quotient rule, the derivative of $ y $ with respect to $ x $ is:

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

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

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

$$ y' = \frac{\sec x \tan x + \sec^3 x \tan x - \sec^3 x}{(1 + \tan x)^2} $$

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

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

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

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

So, the derivative of $ y $ with respect to $ x $ is $ \frac{\sec x \tan x}{(1 + \tan x)^2} $.