- #1
monet A
- 67
- 0
To find dy/dx for
[tex] y = ln(sin^{-1}(x)) [/tex]
I did this:
[tex]
y = ln(sin^{-1}(x))
[/tex]
so
[tex]
e^y = (sin^{-1}(x))
[/tex]
and
[tex]
e^y \frac {dy}{dx} = \frac {1} {\sqrt{1-x^2}}
[/tex]
then
[tex]
\frac {dy}{dx} = \frac {1}{\sqrt {1-x^2} * e^y}
[/tex]
[tex]
= \frac {1}{\sqrt {1-x^2} * e^{ln(sin^{-1}(x))}}
[/tex]
[tex]
= \frac {1}{\sqrt {1-x^2} * (sin^{-1}(x))}}
[/tex]
I think that its all valid implicit differentiation but I m not 100% confident about it, please help.
[tex] y = ln(sin^{-1}(x)) [/tex]
I did this:
[tex]
y = ln(sin^{-1}(x))
[/tex]
so
[tex]
e^y = (sin^{-1}(x))
[/tex]
and
[tex]
e^y \frac {dy}{dx} = \frac {1} {\sqrt{1-x^2}}
[/tex]
then
[tex]
\frac {dy}{dx} = \frac {1}{\sqrt {1-x^2} * e^y}
[/tex]
[tex]
= \frac {1}{\sqrt {1-x^2} * e^{ln(sin^{-1}(x))}}
[/tex]
[tex]
= \frac {1}{\sqrt {1-x^2} * (sin^{-1}(x))}}
[/tex]
I think that its all valid implicit differentiation but I m not 100% confident about it, please help.
Last edited: