Relationship between First and Second Derivatives of a Function

Key Questions

  • Since #f''# is the first derivative of #f'#, #f''# tells us about the increasing/decreasing behavior of #f'#.

  • Answer:

    #2sec^2xtanx#

    Explanation:

    First we find #d/dxtanx#.

    We know that #tanx=sinx/cosx#

    So we can use the quotient rule to solve for this:

    #d/dx(sinx/cosx)=(cosxd/dx(sinx)-sinxd/dx(cosx))/cos^2x#

    #color(white)(d/dx(sinx/cosx))=(cosx(cosx)-sinx(-sinx))/cos^2x#

    #color(white)(d/dx(sinx/cosx))=(cos^2x+sin^2x)/(cos^2x)=1/cos^2x#

    #d/dx(sinx/cosx)=d/dxtanx=sec^2x#

    Now for #d^2/dx^2tanx#, or #d/dxsec^2x#

    Which we can write as #d/dx(secx)^2#, which gives:

    #2secx(secxtanx)#, using the chain rule, where we compute #d/(du)u^2# and #d/dxsecx#.

    Which gives:

    #2sec^2xtanx#

    So:

    #d^2/dx^2tanx=2sec^2xtanx#

  • Answer:

    See below.

    Explanation:

    The second derivative is the derivative of the derivative of a function.

    Let's take a random function, say #f(x)=x^3#. The derivative of #f(x)#, that is, #f'(x)#, is equal to #3x^2#.

    The second derivative of #x^3# is the derivative of #3x^2#. That's #6x#.

    So we say that the second derivative of #f(x)=x^3#, or #f''(x)#, is equal to #6x#

Questions