What is the derivative of (3x - sinx)/cosx?

1 Answer
Jun 4, 2015

The derivative of this can be done with just the quotient rule, but you're free to give your answer simplified in many ways.

d/(dx)([3x - sinx]/(cosx)) = [(cosx*(3-cosx)) - ((3x-sinx)*-sinx)]/(cos^2x)

= [3cosx-cos^2x - (-3xsinx+sin^2x)]/(cos^2x)

= [3cosx-cos^2x + 3xsinx - sin^2x]/(cos^2x)

= 3/(cosx) - 1+ (3xsinx-sin^2x)/(cos^2x)

= 3secx + (3xsinx-sin^2x)(sec^2x) - 1

or

= [3cosx + 3xsinx - (sin^2x + cos^2x)]/(cos^2x)

= [3cosx + 3xsinx - 1]/(cos^2x)

= 3secx + 3xsecxtanx - sec^2x

= - sec^2x + 3secx + 3xsecxtanx

or from Wolfram Alpha:

(3-cosx)secx+(3x-sinx)(secxtanx)