In Exercises 59–86, find the derivative of y with respect to the given independent variable.
69. y = 2^(sin 3t)

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Identify the function to differentiate: $y = 2^{\sin(3t)}$. This is an exponential function where the base is a constant (2) and the exponent is a function of $t$.

Recall the general rule for differentiating functions of the form $a^{u(t)}$, where $a$ is a constant and $u(t)$ is a function of $t$: the derivative is $\frac{dy}{dt} = a^{u(t)} \cdot \ln(a) \cdot \frac{du}{dt}$.

Set $u(t) = \sin(3t)$ and find its derivative $\frac{du}{dt}$. Use the chain rule: the derivative of $\sin(3t)$ is $\cos(3t)$ multiplied by the derivative of \$3t$, which is 3. So, $\frac{du}{dt} = 3 \cos(3t)$.

Apply the formula: $\frac{dy}{dt} = 2^{\sin(3t)} \cdot \ln(2) \cdot 3 \cos(3t)$.

Write the final expression for the derivative as $\frac{dy}{dt} = 3 \ln(2) \cdot 2^{\sin(3t)} \cdot \cos(3t)$, which expresses the rate of change of $y$ with respect to $t$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chain Rule

The chain rule is used to differentiate composite functions. It states that the derivative of a function composed of another function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. For example, if y = f(g(t)), then dy/dt = f'(g(t)) * g'(t).

Derivative of Exponential Functions with Variable Exponents

When differentiating functions like a^(u(t)), where the base a is constant and the exponent u(t) is a function of t, rewrite the function using exponentials and logarithms: a^(u) = e^(u ln a). Then apply the chain rule to differentiate e^(u ln a).

Derivative of Trigonometric Functions

The derivative of sine and cosine functions are fundamental in calculus. Specifically, d/dt[sin(kt)] = k cos(kt), where k is a constant. This rule is essential when differentiating expressions like sin(3t) inside the exponent.

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