Question

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:

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