What is the tangent of y=sqrt(x^2-x+10),\at x=3 ?

The tangent of y=sqrt(x^2-x+10),\at x=3 is y= 5/8 x+17/8

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tangent of y=sqrt(x^2-x+10),\at x=3

tangent of

y = x^{2} - x + 10, at x = 3

y = \frac{5}{8} x + \frac{17}{8}

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

y = x^{2} - x + 10 : \frac{d y}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 10}

Compute the slope of

y = x^{2} - x + 10

x = 3 : m = \frac{5}{8}

Find the line with slope m=

\frac{5}{8}

and passing through

(3, 4) : y = \frac{5}{8} x + \frac{17}{8}

y = \frac{5}{8} x + \frac{17}{8}

\frac{d}{d x} (cos (x) ln (cos (x)) + x sin (x))

\frac{d}{d x} (\frac{ln ( 4 x )}{4 x})

y = 4 - x^{2}, y = x + 2

\frac{\partial}{\partial y} (\frac{1}{x y})

\int \frac{x}{1 + x ^{2}} d x

What is the tangent of y=sqrt(x^2-x+10),\at x=3 ?
The tangent of y=sqrt(x^2-x+10),\at x=3 is y= 5/8 x+17/8