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

Question

Ezekiel Alexander · Accepted Answer

#dy/dx=cosx+sinx#

We need to know the following derivatives:

First we can simplify the function using #tanx=sinx/cosx# and #secx=1/cosx#:

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

Then the derivative is:

#dy/dx=d/dx(sinx)-d/dx(cosx)#

#dy/dx=cosx-(-sinx)#

#dy/dx=cosx+sinx#

Isabella Ackerman · Answer

undefined To find the derivative of $ y = \frac{\tan(x) - 1}{\sec(x)} $, you can use the quotient rule of differentiation, which states that if $ y = \frac{u}{v} $, then $ y' = \frac{u'v - uv'}{v^2} $.

Let $ u = \tan(x) - 1 $ and $ v = \sec(x) $.

Then, differentiate $ u $ and $ v $ separately to find $ u' $ and $ v' $:

$ u' = \sec^2(x) $ (derivative of $ \tan(x) $ is $ \sec^2(x) $)

$ v' = \sec(x) \tan(x) $ (derivative of $ \sec(x) $ is $ \sec(x) \tan(x) $)

Now, plug these values into the quotient rule formula:

$ y' = \frac{(u'v - uv')}{v^2} $

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

$ y' = \frac{(\sec^3(x) - \sec(x)(\tan^2(x) - \tan(x)))}{\sec^4(x)} $

$ y' = \frac{\sec^3(x) - \sec(x)\tan^2(x) + \sec(x)\tan(x)}{\sec^4(x)} $