Overview of Different Functions

Key Questions

  • Answer:

    See below.

    Explanation:

    #d/dxsin^-1x=1/sqrt(1-x^2)#

    #d/dxcos^-1x=-1/sqrt(1-x^2)#

    #tan^-1x=1/(1+x^2)#

    #cot^-1x=-1/sqrt(1+x^2)#

    #sec^-1x=1/(xsqrt(x^2-1))#

    #csc^-1x=-1/(xsqrt(x^2-1))#

    One useful thing to notice is that the derivatives of all inverse "co" functions are equivalent to the derivatives of the original inverse function, but have a negative added.

    Here's a proof for the derivative of the inverse sine function, if you want to avoid memorization. All the other ones can be proved in the same way.

    #y=sin^-1x hArr x=siny#, from the definition of an inverse function.

    Differentiating #x=siny:#

    #d/dx(x)=d/dx(siny)#

    #1=cosy*dy/dx# (Implicit Differentiation)

    #dy/dx=1/cosy#

    We need to get rid of #cosy.# Recall that we said #x=siny# and recall the identity #sin^2theta+cos^2theta=1#. This can be rewritten for #y# and solved for cosine as follows:

    #cos^2y=1-sin^2y#

    #cosy=sqrt(1-sin^2y)#

    Recalling that #x=siny,# then #sin^2y=x^2#

    Thus,

    #cosy=sqrt(1-x^2)#

    #d/dxsin^-1x=1/sqrt(1-x^2)#

    I recommend that you commit these integrals to memory- they will help you when you are learning the trig. substitution.

  • Base e

    #(e^x)'=e^x#

    Other Base

    #(b^x)'=(lnb)b^x#


    I hope that this was helpful.

  • The derivative of a logarithmic function is (1/the function)*derivative of the function.

    For example, #d/dx log x= 1/x#

    Consider another example.

    #d/dx log (1+ x^3)= 1/ (1+x^3) 3x^2#

    In the first example, the function was x. Thus, derivative of the log function was 1/the function *derivative of the function, i.e. , #1/x *1#

    In the second example, the function was #1+x^3#. It's derivative is #3x^2#. Hence derivative of the log function was # 1/ (1+x^3) 3x^2#.

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