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.

Questions