How do you differentiate #y=tlog3e^sint#?

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Answer 1 · 2017-04-26T18:05:32

Please see the explanation.

Given: #y=tlog(3e^sin(t))#

Using the property #log(a^c) = (c)log(x)#

#y = tsin(t)log(3e)#

Differentiate:

#dy/dt = d/dttsin(t)log(3e)#

Use the property that the derivative is a linear operation:

#dy/dt = log(3e)d/dttsin(t)#

#dy/dt = log(3e)(dt/dtsin(t) + td/dtsin(t))#

Simplify:

#dy/dt = log(3e)(sin(t) + tcos(t))#