# How do you differentiate #f(x) = xe^(-2x^2)# using the product rule?

Now that we know what both our derivatives are equal to, let's plug them back into our equation.

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To differentiate the function (f(x) = xe^{-2x^2}) using the product rule, we first identify the two parts of the function that are being multiplied. Here, one part is (u = x) and the other part is (v = e^{-2x^2}). The product rule states that the derivative of a product of two functions (u(x)v(x)) is given by (u'(x)v(x) + u(x)v'(x)).

**Differentiate (u = x):**

[u' = \frac{d}{dx}(x) = 1]

**Differentiate (v = e^{-2x^2}):**

For this part, we use the chain rule, which states that the derivative of a composite function (f(g(x))) is (f'(g(x))g'(x)). Here, (g(x) = -2x^2) and (f(g) = e^g).

[v' = \frac{d}{dx}(e^{-2x^2}) = e^{-2x^2} \cdot \frac{d}{dx}(-2x^2) = e^{-2x^2} \cdot (-4x)]

**Apply the product rule:**

[f'(x) = u'v + uv' = (1)(e^{-2x^2}) + (x)(e^{-2x^2} \cdot (-4x))]

[f'(x) = e^{-2x^2} - 4x^2e^{-2x^2}]

[f'(x) = e^{-2x^2}(1 - 4x^2)]

Therefore, the derivative of (f(x) = xe^{-2x^2}) with respect to (x) is (f'(x) = e^{-2x^2}(1 - 4x^2)).

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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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