Determine the first second and third derivatives of e^x^3

Answer 1

To find the first derivative of \(e^{x^3}\), we can use the chain rule. Let \(u = x^3\), and \(y = e^u\). Then,

\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

We know that \(\frac{dy}{du}\) is simply \(e^u\), and \(\frac{du}{dx}\) is \(3x^2\). Therefore,

\(\frac{dy}{dx} = e^u \cdot 3x^2 = 3x^2e^{x^3}\).

To find the second derivative, we differentiate again:

\(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(3x^2e^{x^3}\right)\).

Using the product rule, we have:

\(\frac{d^2y}{dx^2} = \left(\frac{d}{dx}(3x^2)\right)e^{x^3} + 3x^2\left(\frac{d}{dx}e^{x^3}\right)\).

Differentiating the first term gives us \(6x\), and differentiating \(e^{x^3}\) is \(3x^2e^{x^3}\). Therefore,

\(\frac{d^2y}{dx^2} = 6xe^{x^3} + 3x^2e^{x^3}\).

Finally, to find the third derivative, we differentiate once more:

\(\frac{d^3y}{dx^3} = \frac{d}{dx}\left(6xe^{x^3} + 3x^2e^{x^3}\right)\).

Again, using the product rule, we have:

\(\frac{d^3y}{dx^3} = \left(\frac{d}{dx}(6x)\right)e^{x^3} + 6x\left(\frac{d}{dx}e^{x^3}\right) + \left(\frac{d}{dx}(3x^2)\right)e^{x^3} + 3x^2\left(\frac{d}{dx}e^{x^3}\right)\).

Differentiating \(6x\) gives us \(6\), differentiating \(3x^2\) gives us \(6x\), and differentiating \(e^{x^3}\) is \(3x^2e^{x^3}\). Therefore,

\(\frac{d^3y}{dx^3} = 6e^{x^3} + 6xe^{x^3} + 6xe^{x^3} + 3x^2e^{x^3} = 6e^{x^3} + 12xe^{x^3} + 3x^2e^{x^3}\).

In summary, the derivatives are:

First derivative: \(3x^2e^{x^3}\)

Second derivative: \(6xe^{x^3} + 3x^2e^{x^3}\)

Third derivative: \(6e^{x^3} + 12xe^{x^3} + 3x^2e^{x^3}\)