Suppose u and v are functions of x, and f(x)=ln (u v)+u^{2}. If u(1)=2, v(1)=7, u^{prime}(1)=8, and v^{prime}(1)=21, then f^{prime}(1)=

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Step:1 ：Suppose u and v are functions of x, and f(x)=ln (u v)+u^{2}. If u(1)=2, v(1)=7, u^{prime}(1)=8, and v^{prime}(1)=21, then we are asked to find f^{prime}(1).Step:2 ：The derivative of f(x) can be found using the chain rule and the product rule. The derivative of ln(uv) with respect to x is frac{u&#x27;v+uv&#x27;}{uv} by the chain rule and the product rule. The derivative of u^2 with respect to x is 2uu&#x27; by the chain rule. So the derivative of f(x) is frac{u&#x27;v+uv&#x27;}{uv}+2uu&#x27;.Step:3 ：Substitute the given values of u(1), v(1), u&#x27;(1), and v&#x27;(1) into the derivative of f(x) to find f&#x27;(1).Step:4 ：Final Answer: f^{prime}(1)=boxed{39}

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Step: 1 ：Suppose u and v are functions of x, and f(x)=ln (u v)+u^{2}. If u(1)=2, v(1)=7, u^{prime}(1)=8, and v^{prime}(1)=21, then we are asked to find f^{prime}(1).Step: 2 ：The derivative of f(x) can be found using the chain rule and the product rule. The derivative of ln(uv) with respect to x is frac{u&#x27;v+uv&#x27;}{uv} by the chain rule and the product rule. The derivative of u^2 with respect to x is 2uu&#x27; by the chain rule. So the derivative of f(x) is frac{u&#x27;v+uv&#x27;}{uv}+2uu&#x27;.Step: 3 ：Substitute the given values of u(1), v(1), u&#x27;(1), and v&#x27;(1) into the derivative of f(x) to find f&#x27;(1).Step: 4 ：Final Answer: f^{prime}(1)=boxed{39}

Suppose $u$ and $v$ are functions of $x$ , and $f (x) = \ln (u v) + u^{2}$ . If $u (1) = 2, v (1) = 7, u^{'} (1) = 8$ , and $v^{'} (1) = 21$ , then $f^{'} (1) =$

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Suppose u and v are functions of x, and f(x)=ln⁡(uv)+u2. If u(1)=2,v(1)=7,u′(1)=8, and v′(1)=21, then f′(1)=

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Suppose $u$ and $v$ are functions of $x$ , and $f (x) = \ln (u v) + u^{2}$ . If $u (1) = 2, v (1) = 7, u^{'} (1) = 8$ , and $v^{'} (1) = 21$ , then $f^{'} (1) =$