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,

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

To find the second derivative, we differentiate again:

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

Using the product rule, we have:

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

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

$\frac{d^{2}y}{d{x}^2}=6x{e}^{x^3}+3x^2{e}^{x^3}$.

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

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

Again, using the product rule, we have:

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

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

$\frac{d^{3}y}{d{x}^3}=6e^{x^3}+6x{e}^{x^3}+6x{e}^{x^3}+3x^2{e}^{x^3}=6e^{x^3}+12x{e}^{x^3}+3x^2{e}^{x^3}$.

In summary, the derivatives are:

First derivative: $3x^2{e}^{x^3}$

Second derivative: $6x{e}^{x^3}+3x^2{e}^{x^3}$

Third derivative: $6e^{x^3}+12x{e}^{x^3}+3x^2{e}^{x^3}$