Question

find the slope of the functions graph at the given point. then find an equation for the line tangent to the graph there.
$f(x)=\sqrt{7 x+1},(5,6)$
the slope of the functions graph at $(5,6)$ is □.
(simplify your answer.)

Explanation:

Step1: Use the power rule for differentiation

The function is \( f(x)=\sqrt{7x + 1}=(7x + 1)^{\frac{1}{2}} \).
By the chain - rule \( (u^n)^\prime=nu^{n - 1}\cdot u^\prime \), where \( u = 7x+1\), \(n=\frac{1}{2}\).
First, \(u^\prime=7\). Then \(f^\prime(x)=\frac{1}{2}(7x + 1)^{\frac{1}{2}-1}\cdot7=\frac{7}{2}(7x + 1)^{-\frac{1}{2}}=\frac{7}{2\sqrt{7x + 1}}\).

Step2: Evaluate the derivative at \(x = 5\)

Substitute \(x = 5\) into \(f^\prime(x)\).
When \(x = 5\), \(f^\prime(5)=\frac{7}{2\sqrt{7\times5+1}}=\frac{7}{2\sqrt{35 + 1}}=\frac{7}{2\sqrt{36}}=\frac{7}{2\times6}=\frac{7}{12}\).

Answer:

\(\frac{7}{12}\)