What is the tangent of y=sqrt(x^2-x+5),\at x=5 ?

The tangent of y=sqrt(x^2-x+5),\at x=5 is y= 9/10 x+1/2

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tangent of y=sqrt(x^2-x+5),\at x=5

tangent of

y = x^{2} - x + 5, at x = 5

y = \frac{9}{10} x + \frac{1}{2}

Solution steps

Find the tangent point:

(5, 5)

Find the slope of

y = x^{2} - x + 5 : \frac{d y}{d x} = \frac{2 x - 1}{2 x ^{2} - x + 5}

Compute the slope of

y = x^{2} - x + 5

x = 5 : m = \frac{9}{10}

Find the line with slope m=

\frac{9}{10}

and passing through

(5, 5) : y = \frac{9}{10} x + \frac{1}{2}

y = \frac{9}{10} x + \frac{1}{2}

\int \frac{2 x + 1}{x ^{2} - 2 x + 5} d x

\int csc^{4} (θ) d θ

\frac{d}{d x} (∣ x + 3 ∣)

f^{^{'}} (x) = \frac{1}{3} x^{4} - 5 x^{2} + 5

y = ln (\frac{1}{x})

What is the tangent of y=sqrt(x^2-x+5),\at x=5 ?
The tangent of y=sqrt(x^2-x+5),\at x=5 is y= 9/10 x+1/2