Derivative Rules for y=cos(x) and y=tan(x)

Key Questions

  • What is the derivative of #y=tan(x)# ?

    The derivative of #tanx# is #sec^2x#.

    To see why, you'll need to know a few results. First, you need to know that the derivative of #sinx# is #cosx#. Here's a proof of that result from first principles:

    Once you know this, it also implies that the derivative of #cosx# is #-sinx# (which you'll also need later). You need to know one more thing, which is the Quotient Rule for differentiation:

    Once all those pieces are in place, the differentiation goes as follows:

    #d/dx tanx#
    #=d/dx sinx/cosx#

    #=(cosx . cosx-sinx.(-sinx))/(cos^2x)# (using Quotient Rule)

    #=(cos^2x+sin^2x)/(cos^2x)#

    #=1/(cos^2x)# (using the Pythagorean Identity)

    #=sec^2x#

    Eddie W. · 11 · Aug 18 2014
  • How do you find the derivative of #y=cos(x)# from first principle?

    Using the definition of a derivative:

    #dy/dx = lim_(h->0) (f(x+h)-f(x))/h#, where #h = deltax#

    We substitute in our function to get:

    #lim_(h->0) (cos(x+h)-cos(x))/h#

    Using the Trig identity:

    #cos(a+b) = cosacosb - sinasinb#,

    we get:

    #lim_(h->0) ((cosxcos h - sinxsin h)-cosx)/h#

    Factoring out the #cosx# term, we get:

    #lim_(h->0) (cosx(cos h-1) - sinxsin h)/h#

    This can be split into 2 fractions:

    #lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/h#

    Now comes the more difficult part: recognizing known formulas.

    The 2 which will be useful here are:

    #lim_(x->0) sinx/x = 1#, and #lim_(x->0) (cosx-1)/x = 0#

    Since those identities rely on the variable inside the functions being the same as the one used in the #lim# portion, we can only use these identities on terms using #h#, since that's what our #lim# uses. To work these into our equation, we first need to split our function up a bit more:

    #lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/h#

    becomes:

    #lim_(h->0)cosx((cos h-1)/h) - sinx((sin h)/h)#

    Using the previously recognized formulas, we now have:

    #lim_(h->0) cosx(0) - sinx(1)#

    which equals:

    #lim_(h->0) (-sinx)#

    Since there are no more #h# variables, we can just drop the #lim_(h->0)#, giving us a final answer of: #-sinx#.

    Collin C. · 87 · Aug 22 2014
  • What is the derivative of #y=cos(x)# ?

    #dy/dx=(cos x)'=-sin x#. See derivatives of trig functions for details.

    Calculus V. · · Jul 24 2014

Questions

Differentiating Trigonometric Functions