Let's solve the problem from the image step-by-step. The function given is $f(x) = e^{x^2}$, and we are asked to do the following:

(i) Find $f'(x)$ and $f''(x)$.

1. First Derivative $f'(x)$:

   The function is $f(x) = e^{x^2}$. To differentiate this, we use the chain rule. The derivative of $e^u$ with respect to $u$ is $e^u$, and the derivative of $x^2$ with respect to $x$ is $2x$.

   Therefore, $f'(x) = \frac{d}{dx} \left( e^{x^2} \right) = e^{x^2} \cdot 2x = 2x e^{x^2}.$

2. Second Derivative $f''(x)$:

   Now, to find the second derivative, we differentiate $f'(x) = 2x e^{x^2}$ using the product rule. The product rule states $(uv)' = u'v + uv'$.

   Let $u = 2x$ and $v = e^{x^2}$.

   • $u' = 2$
   • $v' = 2x e^{x^2}$ (from the first derivative we calculated earlier)

   Using the product rule: $f''(x) = \frac{d}{dx}(2x e^{x^2}) = 2 \cdot e^{x^2} + 2x \cdot 2x e^{x^2} = 2e^{x^2} + 4x^2 e^{x^2}.$ Simplifying this: $f''(x) = e^{x^2} (2 + 4x^2).$

(ii) Find the turning point of the curve $y = e^{x^2}$.

A turning point occurs when $f'(x) = 0$. From part (i), we have $f'(x) = 2x e^{x^2}$.

For $f'(x) = 0$, either $2x = 0$ or $e^{x^2} = 0$. Since $e^{x^2}$ is never zero, we have:

$2x = 0 \quad \Rightarrow \quad x = 0.$

So, the turning point occurs at $x = 0$. To find the corresponding $y$-value, we substitute $x = 0$ into $f(x) = e^{x^2}$:

$y = e^{0^2} = e^0 = 1.$

Thus, the turning point is $(0, 1)$.

(iii) Determine whether this turning point is a local maximum or a local minimum.

To classify the turning point, we examine the second derivative $f''(x) = e^{x^2}(2 + 4x^2)$ at $x = 0$:

$f''(0) = e^{0^2}(2 + 4(0)^2) = e^0 \cdot 2 = 2.$

Since $f''(0) > 0$, the function is concave up at $x = 0$, meaning the turning point is a local minimum.

Summary:

1. $f'(x) = 2x e^{x^2}$
2. $f''(x) = e^{x^2}(2 + 4x^2)$
3. The turning point is $(0, 1)$
4. The turning point is a local minimum.

Would you like any further details on these steps, or do you have any specific questions?

Here are some related questions you could explore:

1. How do you find the third derivative of $f(x) = e^{x^2}$?
2. What happens to $f(x)$ as $x \to \infty$?
3. How do you classify points of inflection for $f(x) = e^{x^2}$?
4. Can you find the Taylor series expansion of $f(x) = e^{x^2}$?
5. How does the behavior of the function change for negative values of $x$?

Tip: The second derivative test is a powerful tool to quickly determine the concavity of a function and classify its turning points.