How do you find the derivative of #arcsin e^x#?
1 Answer
Jun 23, 2016
Explanation:
There are two methods:
Using the pre-memorized arcsine derivative:
You may already know that the derivative of arcsine is:
#d/dxarcsin(x)=1/sqrt(1-x^2)#
We can apply the chain rule to this for
#d/dxarcsin(e^x)=1/sqrt(1-(e^x)^2)*d/dxe^x#
#=e^x/sqrt(1-e^(2x))#
Without knowing the arcsine derivative:
Let
#y=arcsin(e^x)#
Thus:
#sin(y)=e^x#
Differentiate both sides (the chain rule will be used on the left-hand side!):
#y^'*cos(y)=e^x#
#y^'=e^x/cos(y)#
Note that we should express
We know that
#sin^2(y)+cos^2(y)=1" "=>" "cos(y)=sqrt(1-sin^2(y))#
#y^'=e^x/sqrt(1-sin^2(y))#
And since
#y^'=e^x/sqrt(1-(e^x)^2)=e^x/sqrt(1-e^(2x))#