Question

find the slope of the graph of the function f(x) = √(2x) at (18,6). then find an equation of the line tangent to the graph at that point. the slope of the graph of the function f(x) = √(2x) at (18,6) is . (type an integer or a simplified fraction.)

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{2x}=(2x)^{\frac{1}{2}}$.

Step2: Find the derivative using the power - rule

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $y=(2x)^{\frac{1}{2}}$, let $u = 2x$, then $y = u^{\frac{1}{2}}$. First, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=2$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. So $f^\prime(x)=\frac{1}{2}(2x)^{-\frac{1}{2}}\cdot2=(2x)^{-\frac{1}{2}}=\frac{1}{\sqrt{2x}}$.

Step3: Evaluate the derivative at the given point

Substitute $x = 18$ into $f^\prime(x)$. $f^\prime(18)=\frac{1}{\sqrt{2\times18}}=\frac{1}{\sqrt{36}}=\frac{1}{6}$.

Answer:

$\frac{1}{6}$