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How do you differentiate #(x+1/x)tanx#?

1 Answer

Feb 20, 2018

#(1-1/x^2)tanx+sec^2x(x+1/x)#

Explanation:

Using product rule
#d/dx(f(x)g(x))=f'(x)g(x)+f(x)g'(x)#
where #f(x)=x+1/x# and #g(x)= tanx#

#(1+(-1)x^(-2))tanx+(x+1/x)sec^2x#
#(1-1/x^2)tanx+(x+1/x)sec^2x#

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