Find y ′ if y = sec − 1 ⁡ ( 3 x ). I know that sec − 1 ⁡ ( x ) = 1 x x 2 − 1 but if I plug in 3 x do I use the product rule on the bottom along with the chain rule?I realize there are easier ways, but I must use Implicit differentiation to solve this problem.

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Find       y    ′   if   y  =      sec          −      1        ⁡  (      3    x    ). I know that       sec          −      1        ⁡  (  x  )  =            1              x                              x            2                    −          1                     but if I plug in       3    x   do I use the product rule on the bottom along with the chain rule?I realize there are easier ways, but I must use Implicit differentiation to solve this problem.

lywyk0 · Accepted Answer

First, this has nothing to do with implicit differentiation; it’s just an ordinary derivative. Secondly,       sec          −      1        ⁡  x is most definitely not       1          x                        x          2                −        1            ; what you mean is that the derivative of       sec          −      1        ⁡  x with respect to   x is       1          x                        x          2                −        1            .Now let’s look at the actual differentiation problem. You have   f  (  x  )  =      sec          −      1        ⁡      3    x  . Differentiating this will require the use of the chain rule, but not the product rule:                                          [                          sec                              −                1                                      ⁡                          3              x                        ]                    ′                                    =                  1                                    3              x                                                                        (                                      3                    x                                    )                                2                            −              1                                      ⋅                              [                          3              x                        ]                    ′                                                  =                  1                                    3              x                                      3              x              −              1                                      ⋅                              [            (            3            x                          )                              1                                  /                                2                                      ]                    ′                                                  =                  1                                    3              x                                      3              x              −              1                                      ⋅                  1          2                (        3        x                  )                      −            1                          /                        2                          ⋅        [        3        x                  ]          ′                                                  =                  1                                    3              x                                      3              x              −              1                                      ⋅                  1          2                ⋅                  1                      3            x                          ⋅        3                                          =                  3                      2                          3              x                                      3              x                                      3              x              −              1                                                                        =                  3                      6            x                          3              x              −              1                                                                        =                  1                      2            x                          3              x              −              1                                          You’re never differentiating a product, so you never use the product rule. (Well, you could use it to differentiate   3  x, but that would be making a great deal of unnecessary work for yourself.)

Finley Mcintosh · Accepted Answer

It might be easier to proceed as follows.  sec  ⁡  (  y  )  =      3    x      ⟹    sec  ⁡  (  y  )  tan  ⁡  (  y  )                    d        y                    d        x              =                    3                    2                  x                    Now plug in for   y in terms of   x, if needed.Hence,                    d        y                    d        x              =                    3                    2                  x                                1              sec        ⁡        (        y        )        tan        ⁡        (        y        )              =                    3                    2                  x                sec        ⁡        (        y        )        tan        ⁡        (        y        )              =                    3                    2                  x                          3          x                                      sec            2                    ⁡          (          y          )          −          1                      =            1              2        x                  3          x          −          1                    For a more &quot;direct&quot; way, using the chain rule, let   z  =      3    x  , then   y  =      sec          −      1        ⁡  (  z  ). Then                    d        y                    d        x              =                    d        y                    d        z                                d        z                    d        x              =            1              z                              z            2                    −          1                                        3                    2                  x                    Now plug in   z  =      3    x   to get the same answer.

Find y′ if y=sec^(−1)(sqrt (3x)).

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