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

Key Questions

  • The derivative of tanxtanx is sec^2xsec2x.

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

    Once you know this, it also implies that the derivative of cosxcosx is -sinxsinx (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 tanxddxtanx
    =d/dx sinx/cosx=ddxsinxcosx

    =(cosx . cosx-sinx.(-sinx))/(cos^2x)=cosx.cosxsinx.(sinx)cos2x (using Quotient Rule)

    =(cos^2x+sin^2x)/(cos^2x)=cos2x+sin2xcos2x

    =1/(cos^2x)=1cos2x (using the Pythagorean Identity)

    =sec^2x=sec2x

  • 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.

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

Questions