What is the tangent of y=sqrt(x^2-x+7),\at x=7 ?

The tangent of y=sqrt(x^2-x+7),\at x=7 is y= 13/14 x+1/2

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tangent of y=sqrt(x^2-x+7),\at x=7

tangent of

y = x^{2} - x + 7, at x = 7

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

Solution steps

Find the tangent point:

(7, 7)

Find the slope of

y = x^{2} - x + 7 : \frac{d y}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 7}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{13}{14}

Find the line with slope m=

\frac{13}{14}

and passing through

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

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

x \to 1 lim (\frac{x ^{9} - 1}{x - 1})

\frac{d}{d x} (tan (\frac{1}{2} x))

\int sin (x + 3) d x

\frac{\partial}{\partial x} (3 x^{2} + 2 y^{2} - 4 x y + 5)

\frac{d}{d t} (\frac{t}{1 + ( 1.5574 + 5.43 t ^{2} ) ^{2}})

What is the tangent of y=sqrt(x^2-x+7),\at x=7 ?
The tangent of y=sqrt(x^2-x+7),\at x=7 is y= 13/14 x+1/2