To find the derivative of the given function, 𝑓(𝑥) = 𝑒^𝑥^2 ⋅ sin(𝑥), we can use the product rule. The product rule states that if we have two functions, 𝑢(𝑥) and 𝑣(𝑥), then the derivative of their product is given by:

𝑑(𝑢(𝑥)⋅𝑣(𝑥))/𝑑𝑥 = 𝑢(𝑥)⋅𝑑𝑣(𝑥)/𝑑𝑥 + 𝑣(𝑥)⋅𝑑𝑢(𝑥)/𝑑𝑥.

In our case, 𝑢(𝑥) = 𝑒^𝑥^2 and 𝑣(𝑥) = sin(𝑥). Let's find the derivatives of these two functions first.

The derivative of 𝑢(𝑥) = 𝑒^𝑥^2 can be found using the chain rule. The chain rule states that if we have a function 𝑔(𝑥) = 𝑓(𝑢(𝑥)), then the derivative of 𝑔(𝑥) with respect to 𝑥 is given by:

𝑑𝑔(𝑥)/𝑑𝑥 = 𝑑𝑓(𝑢(𝑥))/𝑑𝑢(𝑥) ⋅ 𝑑𝑢(𝑥)/𝑑𝑥.

In our case, 𝑔(𝑥) = 𝑒^𝑥^2 and 𝑓(𝑢) = 𝑒^𝑢. So, applying the chain rule, we have:

𝑑𝑔(𝑥)/𝑑𝑥 = 𝑑𝑓(𝑢)/𝑑𝑢 ⋅ 𝑑𝑢(𝑥)/𝑑𝑥 = 𝑒^𝑢 ⋅ 2𝑥.

Since 𝑢 = 𝑥^2, we can substitute 𝑢 in terms of 𝑥:

𝑑𝑔(𝑥)/𝑑𝑥 = 𝑒^(𝑥^2) ⋅ 2𝑥.

Next, let's find the derivative of 𝑣(𝑥) = sin(𝑥). The derivative of sin(𝑥) is cos(𝑥). Therefore:

𝑑𝑣(𝑥)/𝑑𝑥 = cos(𝑥).

Now, let's apply the product rule to find the derivative of 𝑓(𝑥) = 𝑒^𝑥^2 ⋅ sin(𝑥):

𝑑(𝑓(𝑥))/𝑑𝑥 = 𝑢(𝑥)⋅𝑑𝑣(𝑥)/𝑑𝑥 + 𝑣(𝑥)⋅𝑑𝑢(𝑥)/𝑑𝑥 = 𝑒^(𝑥^2)⋅cos(𝑥) + sin(𝑥)⋅2𝑥.

Therefore, the derivative of 𝑓(𝑥) = 𝑒^𝑥^2 ⋅ sin(𝑥) is 𝑒^(𝑥^2)⋅cos(𝑥) + sin(𝑥)⋅2𝑥.

I hope this explanation helps! Let me know if you have any further questions.

References:

- Khan Academy. (n.d.). Product rule. Retrieved from https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-1/v/product-rule