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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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).

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Intro to the Chain Rule

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).

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Derivatives of General Exponential Functions

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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