# How do you find the derivative of #(2e^(3x) + 2e^(-2x))^4#?

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To find the derivative of the given function ((2e^{3x} + 2e^{-2x})^4), you can use the chain rule and the power rule for differentiation.

Let (u = 2e^{3x} + 2e^{-2x}).

Now, differentiate (u) with respect to (x):

[\frac{du}{dx} = \frac{d}{dx}(2e^{3x} + 2e^{-2x})]

Apply the chain rule and the derivative of (e^{ax}) where (a) is a constant:

[\frac{du}{dx} = 2(3e^{3x}) - 2(2e^{-2x})]

Simplify:

[\frac{du}{dx} = 6e^{3x} - 4e^{-2x}]

Now, apply the power rule for differentiation:

[\frac{d}{dx}[(2e^{3x} + 2e^{-2x})^4] = 4(2e^{3x} + 2e^{-2x})^3 \cdot \frac{du}{dx}]

Substitute the expression for (\frac{du}{dx}) we found earlier:

[\frac{d}{dx}[(2e^{3x} + 2e^{-2x})^4] = 4(2e^{3x} + 2e^{-2x})^3 \cdot (6e^{3x} - 4e^{-2x})]

This is the derivative of the given function.

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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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