How to find the derivative of ( 1 - x 2 ) arcsin x ?

Question

How to find the derivative of             (                        1          -                      x            2                              )              arcsin      x      ?

Kathleen Gardner · Accepted Answer

Let,       y    =                  1        -                  x          2                      ⋅    a    r    c          sin      x        .  The desired Derivative can be obtained using the Product Rule for Diffn.Here is, an another way to solve the Problem.We subst.       t    =    a    r    c          sin      x        ,          |      x      |        ≤    1    .    ⇒          sin      t        =    x    ,          |      t      |        ≤          π      2        .  Thus,       y    =                  1        -                              sin            2                    t                      ⋅    t    =    t          cos      t        ,                 where,               t    =          sin      x        ...    .          (      1      )        .  So,       y                 is a fun. of               t    ,                 and, t of               x    .  By the Chain Rule, thus, we have,                    d        y                    d        x              =                  d        y                    d        t              ⋅                  d        t                    d        x              ,        =          {              d                  d          t                            (        t                  cos          t                )            }              {              d                  d          x                            (        a        r        c                  sin          x                )            }        ...    ...          [      ∵      ,              (        1        )            ]        ,        =          {              cos        t            ⋅              d                  d          t                            (        t        )            +      t      ⋅              d                  d          t                            (                  cos          t                )            }              (              1                              1            -                          x              2                                          )        ,        =          {              cos        t            +      t              (        -                  sin          t                )            }              cos      x        ,        =          {                        1          -                      x            2                              -      x      ⋅      a      r      c              sin        x            }              (              1                              1            -                          x              2                                          )        ,        ⇒                  d        y                    d        x              =    1    -                  x        ⋅        a        r        c                  sin          x                                      1          -                      x            2                                ,          |      x      |        &amp;lt;    1    .  N.B.: The Domain of g is [-1,1]; and, that of g&#039; is (-1,1).

How do you find the derivative of (sqrt(1-x^2))arcsin x?

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