What is the tangent of f(x)=(x^2+3)^{1/2},\at x=1 ?

The tangent of f(x)=(x^2+3)^{1/2},\at x=1 is y= 1/2 x+3/2

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tangent of f(x)=(x^2+3)^{1/2},\at x=1

tangent of

f (x) = (x^{2} + 3)^{\frac{1}{2}}, at x = 1

y = \frac{1}{2} x + \frac{3}{2}

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = (x^{2} + 3)^{\frac{1}{2}} : \frac{df ( x )}{d x} = \frac{x}{( x ^{2} + 3 ) ^{\frac{1}{2}}}

Compute the slope of

f (x) = (x^{2} + 3)^{\frac{1}{2}}

x = 1 : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(1, 2) : y = \frac{1}{2} x + \frac{3}{2}

y = \frac{1}{2} x + \frac{3}{2}

\int (1 - \frac{x}{4})^{2} d x

y^{^{'}} = - 2 x y^{3}

x \to \frac{3 π}{2} lim (sec (x))

x \to 0 lim (\frac{x ^{3}}{ln ∣ x ∣})

y^{^{'}} = y (x y^{3} + 5)

What is the tangent of f(x)=(x^2+3)^{1/2},\at x=1 ?
The tangent of f(x)=(x^2+3)^{1/2},\at x=1 is y= 1/2 x+3/2