Question

the equation of the line tangent to the graph of $f(x)=\sqrt{10x}$ at $(10,10)$ is $y=\square$.

Explanation:

Step1: Find the derivative of \(f(x)\)

Rewrite \(f(x)=\sqrt{10x}=(10x)^{\frac{1}{2}}\).
Using the power - rule \((u^n)^\prime = nu^{n - 1}u^\prime\) (where \(u = 10x\), \(n=\frac{1}{2}\)), \(f^\prime(x)=\frac{1}{2}(10x)^{-\frac{1}{2}}\times10=\frac{5}{\sqrt{10x}}\).

Step2: Evaluate the derivative at \(x = 10\)

Substitute \(x = 10\) into \(f^\prime(x)\), \(f^\prime(10)=\frac{5}{\sqrt{10\times10}}=\frac{5}{10}=\frac{1}{2}\). So the slope \(m=\frac{1}{2}\).

Step3: Use the point - slope form \(y - y_1=m(x - x_1)\)

Given the point \((x_1,y_1)=(10,10)\) and \(m=\frac{1}{2}\), \(y - 10=\frac{1}{2}(x - 10)\).
Expand: \(y-10=\frac{1}{2}x - 5\).
Add \(10\) to both sides: \(y=\frac{1}{2}x+5\).

Answer:

\(y=\frac{1}{2}x + 5\)