Question

find the tangent line to ( y = sqrt{x^{2}-x + 7} ) at ( x = 7 ).
the tangent line to ( y = sqrt{x^{2}-x + 7} ) at ( x = 7 ) is
(type an equation.)

Explanation:

Step1: Find the value of \(y\) at \(x = 7\)

Substitute \(x = 7\) into \(y=\sqrt{x^{2}-x + 7}\).
\(y=\sqrt{7^{2}-7 + 7}=\sqrt{49}=7\)

Step2: Differentiate \(y\) using the chain - rule

Let \(u=x^{2}-x + 7\), then \(y = \sqrt{u}=u^{\frac{1}{2}}\).
By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
\(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\) and \(\frac{du}{dx}=2x - 1\).
So \(\frac{dy}{dx}=\frac{2x - 1}{2\sqrt{x^{2}-x + 7}}\)

Step3: Find the slope of the tangent line at \(x = 7\)

Substitute \(x = 7\) into \(\frac{dy}{dx}\).
\(\frac{dy}{dx}\mid_{x = 7}=\frac{2\times7-1}{2\sqrt{7^{2}-7 + 7}}=\frac{14 - 1}{2\times7}=\frac{13}{14}\)

Step4: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=7,y_{1}=7,m=\frac{13}{14}\)
\(y - 7=\frac{13}{14}(x - 7)\)
Expand: \(y-7=\frac{13}{14}x-\frac{13}{2}\)
\(y=\frac{13}{14}x-\frac{13}{2}+7\)
\(y=\frac{13}{14}x-\frac{13}{2}+\frac{14}{2}\)
\(y=\frac{13}{14}x+\frac{1}{2}\)

Answer:

\(y=\frac{13}{14}x+\frac{1}{2}\)