Math Problem: Applying the Product and Chain Rule to Solve for the Derivative of an Integral

ⅆⅆx(∫x2sin(t4)ⅆt)=

To solve this problem, we’ll start by applying the product rule and then the chain rule

To solve this problem, we’ll start by applying the product rule and then the chain rule.

Step 1: Apply the product rule

The derivative of the product of two functions, u(x) and v(x), can be found using the product rule:

d(uv)/dx = u * dv/dx + v * du/dx

In this case, u(x) = ∫x^2 sin(t^4) dt and v(x) = dx/dx = 1 (since dx/dx is simply the derivative of x with respect to x).

So, applying the product rule, we have:

d/dx (∫x^2 sin(t^4) dt) = u * dv/dx + v * du/dx
= (∫x^2 sin(t^4) dt) * 1 + x^2 sin(x^4)

Step 2: Apply the chain rule

Now, we need to apply the chain rule to the first term in the expression obtained in step 1. The integral is with respect to t, but we want to take the derivative with respect to x. Therefore, we need to apply the chain rule.

The chain rule states that if we have a composite function, f(g(x)), then the derivative of f(g(x)) with respect to x is given by:

d(f(g(x)))/dx = f'(g(x)) * g'(x)

In this case, f(g(x)) = ∫x^2 sin(t^4) dt, and we can think of g(x) = x.

Let’s find the derivative of f(g(x)) with respect to x using the chain rule.

f'(g(x)) = d/d(g(x)) (∫x^2 sin(t^4) dt)

Since the integration is with respect to t, we need to use the indefinite integral notation:

f'(g(x)) = d/d(g(x)) (∫x^2 sin(t^4) dt) = ∫x^2 sin(t^4) dt / d(g(x))

Now, to find d(g(x))/dx, we can simply take the derivative of g(x) = x with respect to x:

d(g(x))/dx = 1

Now, we can substitute these values back into the chain rule equation:

d(f(g(x)))/dx = f'(g(x)) * g'(x)
= (∫x^2 sin(t^4) dt) / 1
= ∫x^2 sin(t^4) dt

Therefore, the final answer to the given expression is:

ⅆⅆx(∫x^2 sin(t^4)ⅆt) = ∫x^2 sin(t^4) dt

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