91Ó°ÊÓ

To find the first derivative of \(I(x)^2\), apply the chain rule: \(\frac{\mathrm{d}}{\mathrm{dx}}(I(x)^2) = 2I(x)\frac{\mathrm{dI(x)}}{\mathrm{dx}}\) Now, we need to find \(\frac{\mathrm{dI(x)}}{\mathrm{dx}}\). This is the derivative of the integral with respect to x. Using the Fundamental Theorem of Calculus Part 1, we have: $\frac{\mathrm{dI(x)}}{\mathrm{dx}} = \frac{\mathrm{d}}{\mathrm{dx}} \int_{0}^{x^2} \frac{dt}{\sqrt{1-5t^3}} = \frac{1}{\sqrt{1 - 5x^6}} \cdot \frac{\mathrm{d}(x^2)}{\mathrm{dx}} = 2x\cdot\frac{1}{\sqrt{1-5x^6}}$ Now, we can substitute the result into the earlier expression to find: \(\frac{\mathrm{d}}{\mathrm{dx}}(I(x)^2) = 2I(x)\cdot 2x\cdot \frac{1}{\sqrt{1-5x^6}}\)

To find the second derivative, we apply the product rule to the result from Step 1: \(\frac{\mathrm{d}^{2}}{\mathrm{dx}^{2}}(I(x)^2) = \frac{\mathrm{d}}{\mathrm{dx}}(4xI(x)\cdot \frac{1}{\sqrt{1-5x^6}})\) \( = 4I(x) \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{x}{\sqrt{1-5x^6}}\right) + 4x \cdot \frac{1}{\sqrt{1-5x^6}} \cdot \frac{\mathrm{dI(x)}}{\mathrm{dx}}\) Now, we find the derivative of \(\frac{x}{\sqrt{1-5x^6}}\). Using the quotient rule, we have: \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{x}{\sqrt{1-5x^6}}\right) = \frac{\sqrt{1-5x^6} - x \cdot \frac{1}{2}(1-5x^6)^{-\frac{1}{2}}(-30x^5)}{(1-5x^6)}\) \( = \frac{\sqrt{1-5x^6} + 15x^6}{\sqrt{(1-5x^6)^2}}\) Now, substitute this result back into the expression for the second derivative: \(\frac{\mathrm{d}^{2}}{\mathrm{dx}^{2}}(I(x)^2) = 4I(x) \cdot \frac{\sqrt{1-5x^6} + 15x^6}{\sqrt{(1-5x^6)^2}} + 4x \cdot \frac{1}{\sqrt{1-5x^6}} \cdot 2x\cdot\frac{1}{\sqrt{1-5x^6}}\) This gives us the second derivative of the given function, and our final result is: $\frac{\mathrm{d}^{2}}{\mathrm{dx}^{2}}\left(\int_{0}^{\mathrm{x}^{2}} \frac{\mathrm{dt}}{\sqrt{1-5 \mathrm{t}^{3}}}\right)^{2} = 4I(x) \cdot \frac{\sqrt{1-5x^6} + 15x^6}{\sqrt{(1-5x^6)^2}} + 4x \cdot \frac{1}{\sqrt{1-5x^6}} \cdot 2x\cdot\frac{1}{\sqrt{1-5x^6}}$.