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))/hdydx=limh0f(x+h)f(x)h, where h = deltaxh=δx

    We substitute in our function to get:

    lim_(h->0) (cos(x+h)-cos(x))/hlimh0cos(x+h)cos(x)h

    Using the Trig identity:

    cos(a+b) = cosacosb - sinasinbcos(a+b)=cosacosbsinasinb,

    we get:

    lim_(h->0) ((cosxcos h - sinxsin h)-cosx)/hlimh0(cosxcoshsinxsinh)cosxh

    Factoring out the cosxcosx term, we get:

    lim_(h->0) (cosx(cos h-1) - sinxsin h)/hlimh0cosx(cosh1)sinxsinhh

    This can be split into 2 fractions:

    lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/hlimh0cosx(cosh1)hsinxsinhh

    Now comes the more difficult part: recognizing known formulas.

    The 2 which will be useful here are:

    lim_(x->0) sinx/x = 1limx0sinxx=1, and lim_(x->0) (cosx-1)/x = 0limx0cosx1x=0

    Since those identities rely on the variable inside the functions being the same as the one used in the limlim portion, we can only use these identities on terms using hh, since that's what our limlim 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)/hlimh0cosx(cosh1)hsinxsinhh

    becomes:

    lim_(h->0)cosx((cos h-1)/h) - sinx((sin h)/h)limh0cosx(cosh1h)sinx(sinhh)

    Using the previously recognized formulas, we now have:

    lim_(h->0) cosx(0) - sinx(1)limh0cosx(0)sinx(1)

    which equals:

    lim_(h->0) (-sinx)limh0(sinx)

    Since there are no more hh variables, we can just drop the lim_(h->0)limh0, giving us a final answer of: -sinxsinx.

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

Questions