How do you differentiate #f(x)=(lnx+sinx)(x^2-3x)# using the product rule?
1 Answer
Mar 22, 2017
Explanation:
#"Given " f(x)=g(x)h(x)" then "#
#color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))larr" product rule"#
#color(orange)"Reminder " d/dx(lnx)=1/x ; d/dx(sinx)=cosx#
#"here "g(x)=lnx+sinxrArrg'(x)=1/x+cosx#
#"and " h(x)=x^2-3xrArrh'(x)=2x-3#
#f'(x)=(lnx+sinx)(2x-3)+(x^2-3x)(1/x+cosx)#
