The following differentiated implicitly with respect to θ: 3 x = tan θThe book says 3 d x = sec 2 ⁡ θ d θOne could start the calculation like this (I think): d d θ 3 x = d d θ tan θFrom there I'm not really sure about the steps.

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The following differentiated implicitly with respect to θ:  3  x  =  tan    θThe book says  3  d  x  =      sec    2    ⁡  θ    d  θOne could start the calculation like this (I think):      d          d      θ        3  x  =      d          d      θ        tan    θFrom there I&#039;m not really sure about the steps.

Deja Navarro · Accepted Answer

What is meant by  3  x  =  tan  ⁡  (  θ  )is that both sides are the same function of x or   θ and x and   θ are related somehow. Writing out the relation explicitly we have  3  x  =  tan  ⁡  (  θ  (  x  )  )  .Now define   f  (  x  )  =  3  x and   g  (  x  )  =  tan  ⁡  (  θ  (  x  )  ), this equation means that   f  =  g.Applying the derivative operator to both sides       f    ′    =      g    ′   we have  3  =      θ    ′    (  x  )  sec  ⁡  (  θ  (  x  )      )    2  (by the chain rule and derivative of tan = sec^2). Now you can write it as a differential  3      d    x  =  sec  ⁡  (  θ  (  x  )      )    2        θ    ′    (  x  )      d    x  =  sec  ⁡  (  θ      )    2        d    θsince       θ    ′    (  x  )      d    x  =      d    θ

Nina Bean · Accepted Answer

This is differential notation. If   y  =  f  (  x  ), then it is customary to write   d  y  =      f    ′    (  x  )  d  x. What this says is that a small change in x, dx, produces an approximate change on y of       f    ′    (  x  )  d  x. In your instance you have   x  =  f  (  θ  ) and so   d  x  =      f    ′    (  θ  )  d  θ. The concept comes into play when you talk about linear approximation and you are using the differentials to approximate error. So you have  x  =      1    3    tan  ⁡  θwhich gives  d  x  =      1    3        sec    2    ⁡  (  θ  )  d  θ  .

The following differentiated implicitly with respect to theta : 3x=tan theta

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