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Mathematics 15 Online
OpenStudy (anonymous):

How do I find the second derivative of -1/sin ^2x ?

OpenStudy (anonymous):

is it -1/sin^2 (x)

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

i mean sin square x right?

OpenStudy (anonymous):

write 1 as sin^2x+cos^2x

OpenStudy (anonymous):

it will become-(1+cot^2x)

OpenStudy (anonymous):

u need help?

OpenStudy (anonymous):

actually i was not able to post it on grp chat

OpenStudy (anonymous):

@zain

OpenStudy (anonymous):

I'd just change it to -csc^2(x) and then take the first derivative, and then chain rule for the second derivative

OpenStudy (anonymous):

or else use uv rule

OpenStudy (anonymous):

Thanks guys

OpenStudy (anonymous):

taking 1=u & sinx =v & integrate

OpenStudy (anonymous):

im sry...differentiate , (not integrate)

OpenStudy (anonymous):

Differentiate by parts!! lol :)

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