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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: