Question

question
evaluate \\(\frac{d}{dx} \int_{-2}^{x^4} (t^2 + t)^4 dt\\)

Explanation:

Step1: Apply Leibniz Rule + Chain Rule

Let $u = x^4$, then $\frac{d}{dx}\int_{-2}^{u} (t^2+t)^4 dt = \frac{d}{du}\int_{-2}^{u} (t^2+t)^4 dt \cdot \frac{du}{dx}$

Step2: Use Fundamental Theorem of Calculus

$\frac{d}{du}\int_{-2}^{u} (t^2+t)^4 dt = (u^2+u)^4$

Step3: Compute derivative of $u$

$\frac{du}{dx} = \frac{d}{dx}(x^4) = 4x^3$

Step4: Substitute back $u=x^4$

$(u^2+u)^4 = ((x^4)^2 + x^4)^4 = (x^8 + x^4)^4$

Step5: Combine the terms

Multiply the results from Step2 and Step3: $(x^8 + x^4)^4 \cdot 4x^3$

Answer:

$4x^3(x^8 + x^4)^4$