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

Feb 27, 2016

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

Explanation:

$y = {\left(2 t\right)}^{\sin \left(t\right)}$

$\ln y = \ln \left[{\left(2 t\right)}^{\sin \left(t\right)}\right]$

Use property of logs ...

$\ln y = \sin \left(t\right) \ln \left(2 t\right)$

now use implicit differentiation ...

$\frac{1}{y} \times y ' = \cos \left(t\right) \ln \left(2 t\right) + \sin \frac{t}{2 t} \times 2$

Simplify and solve for $y '$ ...

$y ' = y \left[\cos \left(t\right) \ln \left(2 t\right) + \sin \frac{t}{t}\right]$

Now, insert value for $y$...

$y ' = {\left(2 t\right)}^{\sin \left(t\right)} \left[\cos \left(t\right) \ln \left(2 t\right) + \sin \frac{t}{t}\right]$

hope that helped!