Find the derivative of the function. \[ G(x)=\sqrt{1-49 x^{2}} \arccos (7 x) \] \( G^{\prime}(x)=\square \)

To find the derivative of the function \[ G(x) = \sqrt{1 - 49x^2} \arccos(7x), \] we will use the product rule, which states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product is given by \[ (uv)' = u'v + uv'. \] In this case, we can let \[ u(x) = \sqrt{1 - 49x^2} \quad \text{and} \quad v(x) = \arccos(7x). \] ### Step 1: Find \( u'(x) \) To differentiate \( u(x) \), we use the chain rule. The derivative of \( \sqrt{f(x)} \) is \[ \frac{1}{2\sqrt{f(x)}} f'(x). \] Here, \( f(x) = 1 - 49x^2 \), so \[ f'(x) = -98x. \] Thus, \[ u'(x) = \frac{1}{2\sqrt{1 - 49x^2}} \cdot (-98x) = \frac{-49x}{\sqrt{1 - 49x^2}}. \] ### Step 2: Find \( v'(x) \) To differentiate \( v(x) = \arccos(7x) \), we use the derivative of the arccosine function: \[ \frac{d}{dx} \arccos(g(x)) = -\frac{g'(x)}{\sqrt{1 - (g(x))^2}}. \] Here, \( g(x) = 7x \), so \[ g'(x) = 7. \] Thus, \[ v'(x) = -\frac{7}{\sqrt{1 - (7x)^2}} = -\frac{7}{\sqrt{1 - 49x^2}}. \] ### Step 3: Apply the Product Rule Now we can apply the product rule: \[ G'(x) = u'(x)v(x) + u(x)v'(x). \] Substituting in our expressions for \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ G'(x) = \left(\frac{-49x}{\sqrt{1 - 49x^2}}\right) \arccos(7x) + \sqrt{1 - 49x^2} \left(-\frac{7}{\sqrt{1 - 49x^2}}\right). \] ### Step 4: Simplify The second term simplifies as follows: \[ \sqrt{1 - 49x^2} \left(-\frac{7}{\sqrt{1 - 49x^2}}\right) = -7. \] Thus, we have: \[ G'(x) = \frac{-49x \arccos(7x)}{\sqrt{1 - 49x^2}} - 7. \] ### Final Result The derivative of the function \( G(x) \) is \[ G'(x) = \frac{-49x \arccos(7x)}{\sqrt{1 - 49x^2}} - 7. \]