What is the derivative of #sinx(tanx)#?
1 Answer
Jul 9, 2016
Explanation:
Differentiate using the
#color(blue)"product rule"#
#g(x)=sinxrArrg'(x)=cosx#
#h(x)=tanxrArrh'(x)=sec^2x#
#"--------------------------------------------------"#
Substitute these values into f'(x)
#rArrf'(x)=sinxsec^2x+tanxcosx# Now
#tanxcosx=sinx/cancel(cosx)xxcancel(cosx)=sinx#
#rArrf'(x)=sinxsec^2x+sinx=sinx(sec^2x+1)#