How do you find the derivative of # y = cotx(cscx)#?

1 Answer
Shwetank Mauria
Aug 3, 2016

#(dy)/(dx)=-cscx(csc^2x+cot^2x)#

Explanation:

Product rule states if #f(x)=g(x)h(x)#

then #(df)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)#

Hence as #y=cotx(cscx)#

#(dy)/(dx)=d/(dx)(cotx)xxcscx+d/(dx)(cscx)xxcotx#

= #(-csc^2x)xxcscx+(-cscxcotx)xxcotx#

= #-csc^2x xxcscx-cscxxxcot^2x#

= #-cscx(csc^2x+cot^2x)#

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