Find the derivative of \( y \) with respect \( y=5 x^{2} \arcsin \left(5 x^{2}\right)+\sqrt{1-25 x} \) \( \frac{d y}{d x}=\square \)

We are given \[ y = 5x^2 \arcsin(5x^2) + \sqrt{1-25x}. \] We will differentiate term by term. ### Differentiating \(5x^2 \arcsin(5x^2)\) Let \[ u(x) = 5x^2 \quad \text{and} \quad v(x) = \arcsin(5x^2). \] Using the product rule, \[ \frac{d}{dx}\Bigl[ u(x)v(x) \Bigr] = u'(x)v(x) + u(x)v'(x). \] 1. Differentiate \(u(x)\): \[ u'(x) = \frac{d}{dx}\left(5x^2\right) = 10x. \] 2. Differentiate \(v(x)\) using the chain rule. Let \(w = 5x^2\) so that \[ v(x) = \arcsin(w) \quad \text{and} \quad \frac{dv}{dw} = \frac{1}{\sqrt{1-w^2}}. \] Also, \[ \frac{dw}{dx} = \frac{d}{dx}(5x^2) = 10x. \] Hence, \[ v'(x) = \frac{10x}{\sqrt{1-(5x^2)^2}} = \frac{10x}{\sqrt{1-25x^4}}. \] Thus, the derivative of the first term is \[ \frac{d}{dx}\left[5x^2 \arcsin(5x^2)\right] = 10x\, \arcsin(5x^2) + 5x^2 \cdot \frac{10x}{\sqrt{1-25x^4}} = 10x\, \arcsin(5x^2) + \frac{50x^3}{\sqrt{1-25x^4}}. \] ### Differentiating \(\sqrt{1-25x}\) Rewrite the square root as an exponent: \[ \sqrt{1-25x} = (1-25x)^{\frac{1}{2}}. \] Differentiate using the chain rule: \[ \frac{d}{dx}\left[(1-25x)^{\frac{1}{2}}\right] = \frac{1}{2}(1-25x)^{-\frac{1}{2}} \cdot \frac{d}{dx}(1-25x). \] Since \[ \frac{d}{dx}(1-25x) = -25, \] we have \[ \frac{d}{dx}\left[\sqrt{1-25x}\right] = -\frac{25}{2}(1-25x)^{-\frac{1}{2}} = -\frac{25}{2\sqrt{1-25x}}. \] ### Combining the results The overall derivative is \[ \frac{dy}{dx} = 10x\, \arcsin(5x^2) + \frac{50x^3}{\sqrt{1-25x^4}} - \frac{25}{2\sqrt{1-25x}}. \]