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!

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