find the tangent line to y = \\sqrt{x^{2}-x + 4} at x = 4. the tangent line to y = \\sqrt{x^{2}-x + 4} at x…

Explanation:

Step1: Find the y - value at x = 4

Substitute x = 4 into $y=\sqrt{x^{2}-x + 4}$. $y=\sqrt{4^{2}-4 + 4}=\sqrt{16}=4$

Step2: Differentiate y using the chain - rule

Let $u=x^{2}-x + 4$, 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 $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{2x - 1}{2\sqrt{x^{2}-x + 4}}$.

Step3: Find the slope of the tangent line at x = 4

Substitute x = 4 into $\frac{dy}{dx}$. $\frac{dy}{dx}\big|_{x = 4}=\frac{2\times4-1}{2\sqrt{4^{2}-4 + 4}}=\frac{8 - 1}{2\times4}=\frac{7}{8}$

Step4: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(4,4)$ and $m=\frac{7}{8}$. $y - 4=\frac{7}{8}(x - 4)$ $y-4=\frac{7}{8}x-\frac{7}{2}$ $y=\frac{7}{8}x-\frac{7}{2}+4$ $y=\frac{7}{8}x+\frac{1}{2}$

Answer:

$y=\frac{7}{8}x+\frac{1}{2}$