What is the tangent of f(x)=sqrt(x^2+1),\at x=2 ?

The tangent of f(x)=sqrt(x^2+1),\at x=2 is y= 2/(sqrt(5))x+sqrt(5)-4/(sqrt(5))

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tangent of f(x)=sqrt(x^2+1),\at x=2

tangent of

f (x) = x^{2} + 1, at x = 2

y = \frac{2}{5} x + 5 - \frac{4}{5}

Solution steps

Find the tangent point:

(2, 5)

Find the slope of

f (x) = x^{2} + 1 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 1}

Compute the slope of

f (x) = x^{2} + 1

x = 2 : m = \frac{2}{5}

Find the line with slope m=

\frac{2}{5}

and passing through

(2, 5) : y = \frac{2}{5} x + 5 - \frac{4}{5}

y = \frac{2}{5} x + 5 - \frac{4}{5}

x^{4} - 4 x^{2} + 10, at x = 1

y^{^{'}} + x y = 0

x \to 0 + lim (cot^{2} (x))

\frac{d}{d x} (x^{4} - 9 x + 6)

\frac{d}{d x} (a tan (x))

What is the tangent of f(x)=sqrt(x^2+1),\at x=2 ?
The tangent of f(x)=sqrt(x^2+1),\at x=2 is y= 2/(sqrt(5))x+sqrt(5)-4/(sqrt(5))