Question #eb60c

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Question #eb60c

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Answer 1 · 2017-10-18T21:39:10

$\frac{d y}{d x} = e^{3 x - x^{3}} (3 - 3 x^{2})$ $dy/dx=e^(3x-x^3)(3-3x^2)$

Explanation:

The derivative of $e^{u}$ $e^u$ , where $u$ $u$ is some function of $x$ $x$ , is $\frac{d y}{d x} = e^{u} \cdot \frac{d u}{d x}$ $dy/dx=e^u*(du)/dx$ .

So, in this case, the derivative is:

$\frac{d y}{d x} = e^{3 x - x^{3}} \cdot \frac{d}{d x} (3 x - x^{3})$ $dy/dx=e^(3x-x^3)*d/dx(3x-x^3)$

Simplifying gives:

$\frac{d y}{d x} = e^{3 x - x^{3}} (3 - 3 x^{2})$ $dy/dx=e^(3x-x^3)(3-3x^2)$

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