Home > Calculus > Derivatives of y=sec(x), y=cot(x), y= csc(x)

How do you find the derivative of #cscx#?

Answer 1

Christopher Allers

#(dy)/(dx)=-cotxcscx#

Rewrite #""cscx""# in terms of #""sinx""# and use the quotient rule

quotient rule #" "y=u/v=>(dy)/(dx)=(vu'-uv')/v^2#

#y=cscx=1/sinx#

#u=1=>u'=0#

#v=sinx=>v'=cosx#

#(dy)/(dx)=((sinx xx0)-(1xxcosx))/(sinx)^2#

#(dy)/(dx)=(0-cosx)/(sinx)^2#

#(dy)/(dx)=-cosx/(sinxsinx)=-cosx/sinx xx 1/sinx#

#(dy)/(dx)=-cotxcscx#

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Answer 2

Scarlett Allison

To find the derivative of csc(x), you can use the quotient rule. The derivative of csc(x) with respect to x is equal to negative cosec(x) cot(x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.