How do you differentiate #f(x)= ( x + 1 )/ ( csc x )# using the quotient rule?

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Answer 1 · 2016-01-15T02:42:54

#f'(x)=((x+1)cotx+1)/cscx#

The quotient rule states that

#d/dx[(g(x))/(h(x))]=(g'(x)h(x)-h'(x)g(x))/[g(x)]^2#

Applying this to the function at hand, we see that

#f'(x)=(cscxd/dx[x+1]-(x+1)d/dx[cscx])/csc^2x#

Which gives

#f'(x)=((cscx)(1)-(x+1)(-cscxcotx))/csc^2x#

Simplify.

#f'(x)=(cscx(1+(x+1)cotx))/csc^2x#

#f'(x)=((x+1)cotx+1)/cscx#