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

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

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

tangent of

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

y = \frac{1}{2} x + 2 - \frac{1}{2}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

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{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(1, 2) : y = \frac{1}{2} x + 2 - \frac{1}{2}

y = \frac{1}{2} x + 2 - \frac{1}{2}

x \to 2 lim (3 x^{2} - 5 x + 1)

\int_{0}^{1} x^{5} d x

x^{3} y^{^{'''}} - 3 x^{2} y^{^{''}} + 6 x y^{^{'}} - 6 y = 0

15 t^{2} + 9

\frac{\partial}{\partial x} (7 x^{2} e^{y})

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