Overview of Different Functions
Key Questions

Answer:
See below.
Explanation:
#d/dxsin^1x=1/sqrt(1x^2)# #d/dxcos^1x=1/sqrt(1x^2)# #tan^1x=1/(1+x^2)# #cot^1x=1/sqrt(1+x^2)# #sec^1x=1/(xsqrt(x^21))# #csc^1x=1/(xsqrt(x^21))# 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=1sin^2y# #cosy=sqrt(1sin^2y)# Recalling that
#x=siny,# then#sin^2y=x^2# Thus,
#cosy=sqrt(1x^2)# #d/dxsin^1x=1/sqrt(1x^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# .