How do you find the derivative of #sinx(tanx)#?

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Answer 1 · 2018-03-29T08:34:48

#d/dx(sinxtanx)=cosxtanx+sinxsec^2x#

After simplification #->sinx+tanxsecx#

#(uv)'=u'v+uv'#

#u=sinx, v=tanx#

Therefore

#d/dx(sinxtanx)=(d/dxsinx)tanx+sinx(d/dxtanx)#

#=cosxtanx+sinxsec^2x#

We could simplify this answer a bit by using some basic trig identities:

#=cancelcosx(sinx/cancelcosx)+sinx(1/cos^2x)#

#=sinx+sinx/cosx(1/cosx)#

#=sinx+tanxsecx#