if ( y = 5xsqrt{x^{2}+1} ), then ( \frac{dy}{dx} ) at ( x = 3 ) is

Explanation:

Step1: Apply the product rule

The product rule states that if (y = uv), then (y^\prime=u^\prime v + uv^\prime). Let (u = 5x) and (v=\sqrt{x^{2}+1}=(x^{2}+1)^{\frac{1}{2}}). (u^\prime = 5). To find (v^\prime), use the chain rule. Let (t=x^{2}+1), then (v = t^{\frac{1}{2}}). (v^\prime=\frac{dv}{dt}\cdot\frac{dt}{dx}). (\frac{dv}{dt}=\frac{1}{2}t^{-\frac{1}{2}}) and (\frac{dt}{dx}=2x). So (v^\prime=\frac{1}{2}(x^{2}+1)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+1}}).

Step2: Calculate (y^\prime)

By the product rule (y^\prime=5\sqrt{x^{2}+1}+5x\cdot\frac{x}{\sqrt{x^{2}+1}}). Simplify (y^\prime=\frac{5(x^{2}+1)+5x^{2}}{\sqrt{x^{2}+1}}=\frac{5x^{2}+5 + 5x^{2}}{\sqrt{x^{2}+1}}=\frac{10x^{2}+5}{\sqrt{x^{2}+1}}).

Step3: Substitute (x = 3)

When (x = 3), (y^\prime=\frac{10\times3^{2}+5}{\sqrt{3^{2}+1}}). (y^\prime=\frac{90 + 5}{\sqrt{10}}=\frac{95}{\sqrt{10}}=\frac{95\sqrt{10}}{10}=\frac{19\sqrt{10}}{2}).

Answer:

(\frac{19\sqrt{10}}{2})