Optimizing Your Math Skills | Applying the Chain Rule to Solve Complex Derivatives

ⅆⅆx(∫x2sin(t4)ⅆt)=

To solve this problem, we can start by applying the chain rule of derivatives

To solve this problem, we can start by applying the chain rule of derivatives.

Let’s break down the problem step by step:

Step 1: Begin with the given expression:
d/dx(∫x^2sin(t^4)dt)

Step 2: Apply the chain rule. We need to differentiate the integrand with respect to its variable t and then multiply by the derivative of the inside function, which in this case is x^2 with respect to x.

d/dx(∫x^2sin(t^4)dt) = (∂/∂x(∫x^2sin(t^4)dt)) * (∂x^2/∂x)

Step 3: Simplify the second term, (∂x^2/∂x):

The derivative of x^2 with respect to x is simply 2x. So, (∂x^2/∂x) = 2x.

d/dx(∫x^2sin(t^4)dt) = (∂/∂x(∫x^2sin(t^4)dt)) * 2x

Step 4: Differentiate the integral term (∂/∂x(∫x^2sin(t^4)dt)):

To differentiate the integral term with respect to x, we treat x as a constant inside the integral and differentiate with respect to t, since the integral is with respect to t.

d/dx(∫x^2sin(t^4)dt) = (∂/∂x(∫x^2sin(t^4)dt)) * 2x
= (∫∂(x^2sin(t^4))/∂x * dt) * 2x
= (∫2xsin(t^4)dt) * 2x

Step 5: Simplify the integral term (∫2xsin(t^4)dt):

Integrating 2x with respect to t will simply give us 2xt. Therefore, the integral of 2xsin(t^4)dt is 2xt + C, where C is the constant of integration.

d/dx(∫x^2sin(t^4)dt) = (∫2xsin(t^4)dt) * 2x
= (2xt + C) * 2x
= 4x^2t + 2Cx

So, the final answer is 4x^2t + 2Cx.

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