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…

Explanation:

Step1: Find the y - value 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 $y = \sqrt{u}=u^{\frac{1}{2}}$. First, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$, and $\frac{du}{dx}=2x - 1$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{2x - 1}{2\sqrt{x^{2}-x + 7}}$.

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

Substitute x = 7 into $\frac{dy}{dx}$. $\frac{dy}{dx}\big|_{x = 7}=\frac{2\times7-1}{2\sqrt{7^{2}-7 + 7}}=\frac{14 - 1}{2\times7}=\frac{13}{14}$

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(7,7)$ and $m=\frac{13}{14}$. $y - 7=\frac{13}{14}(x - 7)$ $y-7=\frac{13}{14}x-\frac{13}{2}$ $y=\frac{13}{14}x-\frac{13}{2}+7$ $y=\frac{13}{14}x+\frac{1}{2}$

Answer:

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