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

Please see the explanation.

Differentiate:

Use the property that the derivative is a linear operation:

Apply the product rule:

Simplify:

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To differentiate ( y = t \log_{3} (e^{\sin t}) ), you can use the product rule and the chain rule. First, rewrite the function as ( y = t \log_{3} (e^{\sin t}) ) = ( y = t \ln (e^{\sin t}) / \ln(3) ).

Now, differentiate using the product rule and chain rule:

( y' = \frac{d}{dt} [t] \cdot \frac{1}{\ln(3)} \ln(e^{\sin t}) + t \cdot \frac{d}{dt} [\frac{1}{\ln(3)} \ln(e^{\sin t})] )

( y' = 1/\ln(3) \cdot \ln(e^{\sin t}) + t \cdot \frac{1}{\ln(3)} \cdot \frac{d}{dt} [\ln(e^{\sin t})] )

( y' = 1/\ln(3) \cdot \ln(e^{\sin t}) + t \cdot \frac{1}{\ln(3)} \cdot \cos(t) )

( y' = 1/\ln(3) \cdot \sin(t) + t \cdot \frac{\cos(t)}{\ln(3)} )

( y' = \frac{\sin(t)}{\ln(3)} + \frac{t \cos(t)}{\ln(3)} )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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