How do you differentiate #f(x)=csc(sqrt(x^2-5x)) # using the chain rule?

1 Answer

#f'(x)=-(2x-5)/(2sqrt (x^2-5x))csc (sqrt (x^2-5x))cot (sqrt (x^2-5x))#

Explanation:

#f (x)=csc(sqrt (x^2-5x))#
Differentiating both sides with respect to #x# we get
#f'(x)=-csc (sqrt (x^2-5x)).cot (sqrt (x^2-5x))×d/dx (sqrt (x^2-5x))#
#=>f'(x)=-csc (sqrt (x^2-5x)).cot (sqrt (x^2-5x))×1/(2sqrt (x^2-5x))×(2x-5)#
#:.f'(x)=-(2x-5)/(2sqrt (x^2-5x))csc (sqrt (x^2-5x))cot (sqrt (x^2-5x))#