How to find d y d x by implicit differentiation of tan ( x + y ) = x and evaluate at point (0,0)?

Question

How to find                     d        y                    d        x             by implicit differentiation of             tan              (        x        +        y        )              =    x   and evaluate at point (0,0)?

nebnikwjq · Accepted Answer

The following crucial steps must be taken when performing implicit differentiation:Take the derivative of both sides of the equation with respect to       x  .Differentiate terms with       x   as normal. Differentiate terms with       y   as normal too but tag on a                     d        y                    d        x             to the end.Solve for the                     d        y                    d        x            . Let&#039;s contrast the two sides now:            d              d        x                    [              tan                  (          x          +          y          )                    ]        =          d              d        x                    [      x      ]      The right hand side just comes out as       1  , but the left hand side will require that we use a chain rule. This goes as:            d              d        x                    [              tan                  (          x          +          y          )                    ]        ⋅          d              d        x                    [      x      +      y      ]      This comes out to:                    sec        2                    (        x        +        y        )              ⋅          (      1      +                        d          y                          d          x                    )      Putting this back in the whole equation:                    sec        2                    (        x        +        y        )              +                  sec        2                    (        x        +        y        )                            d        y                    d        x              =    1  Now, you just solve for                     d        y                    d        x             using some basic algebra:                    d        y                    d        x              =                  1        -                              sec            2                                (            x            +            y            )                                                sec          2                          (          x          +          y          )                    Now, you&#039;re given the point             (      0      ,      0      )       to evaluate the derivative at. All you do is plug this in:                    d        y                    d        x              =                  1        -                              sec            2                                (            0            )                                                sec          2                          (          0          )                                sec              (        0        )             is simply             1              cos                  (          0          )                    . Since             cos              (        0        )             = 1,             sec              (        0        )             is also 1. So, wherever we see             sec              (        0        )            , we just plug in       1  :                    d        y                    d        x              =                  1        -        1            1        =    0  So your tangent line would have a slope of       0   at the point             (      0      ,      0      )      .

jdkhoo88o8j · Accepted Answer

Chain rule differentiation is just another name for implicit differentiation.                    d                  tan                      (            x            +            y            )                                      d                  (          x          +          y          )                      ×                  d                  (          x          +          y          )                            d        x            =                    d        x                    d        x                                sec        2                    (        x        +        y        )              ×          (      1      +                        d          y                          d          x                    )        =    1  Rearranging:      1    +                  d        y                    d        x              =                  cos        2                    (        x        +        y        )                                d        y                    d        x              =                  cos        2                    (        x        +        y        )              -    1  Substituting             (      0      ,      0      )       in above equation,we get:                    d        y                    d        x              =                  cos        2                    (        0        )              -    1  i.e                     d        y                    d        x              =    0

How to find dy/dx by implicit differentiation of tan(x+y)=x and evaluate at point (0,0)?

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