Overview of Different Functions
Key Questions
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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 saidx=siny and recall the identitysin^2theta+cos^2theta=1 . This can be rewritten fory and solved for cosine as follows:cos^2y=1-sin^2y cosy=sqrt(1-sin^2y) Recalling that
x=siny, thensin^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.
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Base e
(e^x)'=e^x Other Base
(b^x)'=(lnb)b^x
I hope that this was helpful.
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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 is3x^2 . Hence derivative of the log function was1/ (1+x^3) 3x^2 .