Determine the first second and third derivatives of e^x^3
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: \(6e^{x^3} + 12xe^{x^3} + 3x^2e^{x^3}\)
To find the first, second, and third derivatives of the function f(x) = e^(x^3), we'll use the chain rule repeatedly.
First, let's find the first derivative (f'(x)):
f'(x) = d/dx(e^(x^3))
To apply the chain rule, consider the function g(u) = e^u, where u = x^3. The derivative of g(u) with respect to u is simply g'(u) = e^u.
Using the chain rule, we have:
f'(x) = d/dx(e^(x^3)) = g'(u) * d/dx(x^3) = e^(x^3) * d/dx(x^3)
To find d/dx(x^3), we differentiate x^3 with respect to x:
d/dx(x^3) = 3x^2
Substituting back into the equation, we get:
f'(x) = e^(x^3) * 3x^2
Next, let's find the second derivative (f''(x)):
f''(x) = d/dx(f'(x))
To find this, we need to differentiate f'(x) = e^(x^3) * 3x^2 using the product rule.
Using the product rule, we have:
f''(x) = (e^(x^3))' * 3x^2 + e^(x^3) * (3x^2)'
The derivative of e^(x^3) with respect to x is simply the derivative of the exponent multiplied by e^(x^3):
(e^(x^3))' = (x^3)' * e^(x^3) = 3x^2 * e^(x^3)
The derivative of 3x^2 with respect to x is simply 6x.
Substituting these derivatives back into the equation, we get:
f''(x) = 3x^2 * e^(x^3) + e^(x^3) * 6x
Finally, let's find the third derivative (f'''(x)):
f'''(x) = d/dx(f''(x))
To find this, we need to differentiate f''(x) = 3x^2 * e^(x^3) + e^(x^3) * 6x.
Using the product rule, we differentiate each term separately:
f'''(x) = (3x^2 * e^(x^3) + e^(x^3) * 6x)' = (6x * e^(x^3) + 3x^2 * e^(x^3)) + (e^(x^3) * 6)
Simplifying, we have:
f'''(x) = 6x * e^(x^3) + 3x^2 * e^(x^3) + 6 * e^(x^3)
So, the first, second, and third derivatives of f(x) = e^(x^3) are:
f'(x) = e^(x^3) * 3x^2
f''(x) = 3x^2 * e^(x^3) + e^(x^3) * 6x
f'''(x) = 6x * e^(x^3) + 3x^2 * e^(x^3) + 6 * e^(x^3)