How do you find the derivative of #lnsin^-1(x)#?

2 Answers
Anjali G
Nov 6, 2016

#d/dx(lnsin^-1x) = 1/((sin^-1x)sqrt(1-x^2))#

Explanation:

Use the derivative of natural logs rule:
#d/dx(lnu)=(u')/u#

So, find the derivative of #sin^-1x#, and divide it by #sin^-1x#.
The derivative of #sin^-1x# is #1/(sqrt(1-x^2))#
#d/dx(lnsin^-1x) = (1/(sqrt(1-x^2)))/(sin^-1x)#

After simplifying, we get
#d/dx(lnsin^-1x) = 1/((sin^-1x)sqrt(1-x^2))#

Matthew T.
Nov 6, 2016

#\frac1{sin^-1(x)}\cdot\frac1\sqrt{1-x^2}#

Explanation:

This question is testing your knowledge of the chain rule.
The chain rule states that the derivative of a function#f(u)# where #u# is another function is#f'(u)\cdot\frac{du}dx#, so the derivative of #ln(sin^-1(x))# is #\frac1{sin^-1(x)}\cdot\fracddx[sin^-1(x)]#
And because #\fracddx[sin^-1(x)]=\frac1\sqrt{1-x^2}#,
#\fracddx[ln(sin^-1(x))]=\frac1{sin^-1(x)}\cdot\frac1\sqrt{1-x^2}#

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