How to find the derivative of y = tan x + cot x ?

Question

How to find the derivative of       y    =          tan      x        +          cot      x      ?

Kylie Woodward · Accepted Answer

Explanation: Separately address the tanx and the cotx. Don't forget about trig function derivatives:

\frac{d}{d x} (\sin (x)) = \cos (x) \frac{d}{d x} (\tan (x)) = \sec^{2} (x) \frac{d}{d x} (\sec (x)) = \sec (x) \tan (x) \frac{d}{d x} (\cos (x)) = - \sin (x) \frac{d}{d x} (\cot (x)) = - \csc^{2} (x) \frac{d}{d x} (\csc (x)) = - \csc (x) \cot (x)

Derivative of tan x is

\sec^{2} x

. Derivative of cot x is

- \csc^{2} x

. Since they are adding together, we can treat them it as tan'x and cot'(x):

f' (x) = \sec^{2} (x) - \csc^{2} (x)

tammypierce5kgk · Accepted Answer

Solution:A slightly different approach...Rewrite in terms of sine and cosine.        y    =                  sin        x                    cos        x              +                  cos        x                    sin        x                  y    =                                        sin            2                    x                +                              cos            2                    x                                      sin          x                          cos          x                          y    =          1                        sin          x                          cos          x                          y    =                  (                  sin          x                          cos          x                )                    -        1            Step 2To differentiate, apply the chain rule. Let       y    =          u              -        1             and       u    =          sin      x              cos      x      . The product rule can be used to differentiate the function u.                    d        u                    d        x              =          cos      x              (              cos        x            )        +          sin      x              (      -              sin        x            )        =                  cos        2            x        -                  sin        2            x        =          cos      2        x   By the power rule,                     d        y                    d        u              =    -          1              u        2                  y    ′    =                  d        y                    d        u              ⋅                  d        u                    d        x                  y    ′    =          cos      2        x    ⋅    -          1              u        2                  y    ′    =    -                  cos                  (          2          x          )                                      (                      sin            x                                cos            x                    )                2            Answer is       y    ′    =    -          cos      2        x                  sec        2            x                      csc        2            x

How to find the derivative of y = tan x + cot x ?

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