find the tangent line to ( y = sqrt{x^{2}-x + 7} ) at ( x = 7 ).\nthe tangent line to ( y = sqrt{x^{2}-x +…

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})