How do you find the derivative of #g(t)=(2t)^(sin(t))#?

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Answer 1 · 2016-02-27T01:01:58

Take the natural log of both sides, then use implicit differentiation ...

#y=(2t)^(sin(t))#

#lny=ln[(2t)^(sin(t))]#

Use property of logs ...

#lny=sin(t)ln(2t)#

now use implicit differentiation ...

#1/yxxy'=cos(t)ln(2t)+sint/(2t)xx2#

Simplify and solve for #y'# ...

#y'=y[cos(t)ln(2t)+sint/(t)]#

Now, insert value for #y#...

#y'=(2t)^(sin(t))[cos(t)ln(2t)+sint/(t)]#

hope that helped!