If \(y=5x\sqrt{x^2+1}\), then \(\dfrac{dy}{dx}\) at \(x=3\) is

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$$ y=5x\sqrt{x^2+1}$$ $$ y=5[(x)(x^2+1)^{\frac{1}{2}}] $$ $$ \frac{dy}{dx}=5\left[(x)\left(\frac{1}{2}\cdot \frac{1}{\sqrt{x^2+1}}\cdot 2x\right)+(\sqrt{x^2+1})(1)\right] $$ $$ \frac{dy}{dx} = \frac{5x^2}{\sqrt{x^2+1}}+5\sqrt{x^2+1} $$ $$ \frac{dy}{dx}\Big|_{x=3} = \frac{5(3)^2}{\sqrt{3^2+1}}+5\sqrt{3^2+1} $$ $$ = \boxed{\frac{45}{\sqrt{10}}+5\sqrt{10}} $$