What is derivative of function y=1/(sin x)^2?



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Posted on (Answer #1)

You need to differentiate the function with respect to x, hence, you need to use the quotient rule and the chain rule, such that:

`y' = (1'*(sin^2 x) - 1*(sin^2 x)')/(sin^4 x)`

`y' = (0*(sin^2 x) - 1*2sin x*(cos x))/(sin^4 x)`

`y' = -(2sin x*cos x)/(sin^4 x)`

Using the double angle identity `2sin x*cos x = sin 2x` , yields:

`y' = -(sin 2x)/(sin^4 x)`

You also may reduce duplicate factors, such that:

`y' = -(2cos x)/(sin^3 x)`

Hence, evaluating the derivative of the given function yields `y' = -(sin 2x)/(sin^4 x).`

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