How do you differentiate #f(x)=xcscx# using the product rule?

1 Answer
Dec 3, 2015

Answer:

#f'(x)=cscx (1- xcotx)#

Explanation:

Note: #F(x) = f(x)*g(x)#
Then derivative : #F'(x) = f'(x)*g(x) + g'(x)*f(x)#

Note: #f(x) = csc x hArr f'(x)= -csc x cot x#

==================

given #f(x) = x csc x#
#f'(x) = d/dx (x) * csc x + x d/dx cscx#

# = 1* csc x + x*(-csc x cotx )#
#= csc x - x cscx cot x#
#color(red)(=cscx (1- xcotx)#

========================
FYI: How to derive derivative of #f(x) = csc x #
#f(x) = 1/(sinx)#

(using quotient rule)

#f'(x) = (-cos x)/(sin^2x)#

#f'(x) = (1/sinx)*(-cos x/sinx)#

#f'(x)= -csc x cot x#