How do you find the derivative of lnsin^-1(x)lnsin1(x)?

2 Answers
Nov 6, 2016

d/dx(lnsin^-1x) = 1/((sin^-1x)sqrt(1-x^2))ddx(lnsin1x)=1(sin1x)1x2

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))

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 functionf(u) where u is another function isf'(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}