Finding derivative f ( x ) = 2 x 3 I have to find the derivative and the slope at a=6The function is f ( x ) = 2 x 3 I have to find the answer using the formula, f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x I tried getting rid of the denominator, but I think I'm getting mixed up somewhere.The answer book says the slope is − 1 216 Here's what I've done, f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = lim Δ x → 0 2 ( x + Δ x ) 3 − 2 x 3 Δ x = lim Δ x → 0 ( 2 ( x + Δ x ) 3 − 2 x 3 ) ( ( x + Δ x ) 3 ( x 3 ) ) Δ x ( ( x + Δ x ) 3 ( x 3 ) ) = lim Δ x → 0 2 ( x 3 ) − 2 ( x + Δ x ) 3 Δ x ( x + Δ x ) 3 ( x 3 ) To cancel out the denominator.Did I do this right??Should I now expand the parentheses and cancel things out??Would I still get the same answer as the answer book??(Because the answer book have made a typo once before).Thanks

Question

Finding derivative   f  (  x  )  =            2              x        3            I have to find the derivative and the slope at a=6The function is   f  (  x  )  =            2              x        3            I have to find the answer using the formula,      f    ′    (  x  )  =      lim          Δ      x      →      0                          f        (        x        +        Δ        x        )        −        f        (        x        )                    Δ        x            I tried getting rid of the denominator, but I think I&#039;m getting mixed up somewhere.The answer book says the slope is                     −        1            216      Here&#039;s what I&#039;ve done,      f    ′    (  x  )  =      lim          Δ      x      →      0                          f        (        x        +        Δ        x        )        −        f        (        x        )                    Δ        x              =      lim          Δ      x      →      0                                                2                          (              x              +              Δ              x                              )                3                                                    −                              2                          x              3                                                  Δ        x                =      lim          Δ      x      →      0                          (                              2                          (              x              +              Δ              x                              )                3                                                    −                              2                          x              3                                      )        (        (        x        +        Δ        x                  )          3                (                  x          3                )        )                    Δ        x        (        (        x        +        Δ        x                  )          3                (                  x          3                )        )                  =      lim          Δ      x      →      0                          2        (                  x          3                )        −        2        (        x        +        Δ        x                  )          3                            Δ        x        (        x        +        Δ        x                  )          3                (                  x          3                )            To cancel out the denominator.Did I do this right??Should I now expand the parentheses and cancel things out??Would I still get the same answer as the answer book??(Because the answer book have made a typo once before).Thanks

Myla Pierce · Accepted Answer

You&#039;re absolutely right!                              f          ′                (        x        )                            =                  lim                      Δ            →            0                                                2                          x              3                        −            2            (                          x              3                        +            3                          x              2                        Δ            x            +            3            x                          Δ              2                        x            +                          Δ              3                        x            )                                Δ            x            (            x            +            Δ            x                          )              3                        (                          x              3                        )                                                            =                  lim                      Δ            x            →            0                                                −            6                          x              2                        −            6            x            Δ            x            −            2                          Δ              3                        x                                (            x            +            Δ            x                          )              3                        (                          x              3                        )                                                            =                  lim                      Δ            x            →            0                                                −                          x              2                                                      (                                      6            +            6                                          Δ                x                            x                        +            2                                                            Δ                  3                                x                                            x                2                                                                    )                                                          (            x            +            Δ            x                          )              3                        (                          x              3                        )                                                            =                  lim                      Δ            x            →            0                                                −                                          (                                      6            +            6                                          Δ                x                            x                        +            2                                                            Δ                  3                                x                                            x                2                                                                    )                                                          (            x            +            Δ            x                          )              3                        x                                                            =        −                  6                      x            4

Reginald Delacruz · Accepted Answer

Yes, so far this is correct. Now you should expand the parenthesis and cancel the   2      x    3   first. Then you can cancel the   Δ  x. You should then be able to get the answer from the book.

Finding derivative f ( x ) = <mrow class="MJX-TeXAtom-ORD"> 2

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