What is the tangent of f(x)=sqrt(x+2) ?

The tangent of f(x)=sqrt(x+2) is y= 1/(2sqrt(a_{0)+2)}x+sqrt(a_{0)+2}-(a_{0})/(2sqrt(a_{0)+2)}

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tangent of f(x)=sqrt(x+2)

tangent of

f (x) = x + 2

y = \frac{1}{2 a _{0} + 2} x + a_{0} + 2 - \frac{a _{0}}{2 a _{0} + 2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} + 2)

Find the slope of

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

Compute the slope of

f (x) = x + 2

x = a_{0} : m = \frac{1}{2 a _{0} + 2}

Find the line with slope m=

\frac{1}{2 a _{0} + 2}

and passing through

(a_{0}, a_{0} + 2) : y = \frac{1}{2 a _{0} + 2} x + a_{0} + 2 - \frac{a _{0}}{2 a _{0} + 2}

y = \frac{1}{2 a _{0} + 2} x + a_{0} + 2 - \frac{a _{0}}{2 a _{0} + 2}

\frac{\partial}{\partial x} (3 y^{2} + x^{2} - 3)

y^{^{'}} = cos (2 x)

25 cos (\frac{π x}{50})

\int cos^{3} (a x + b) d x

What is the tangent of f(x)=sqrt(x+2) ?
The tangent of f(x)=sqrt(x+2) is y= 1/(2sqrt(a_{0)+2)}x+sqrt(a_{0)+2}-(a_{0})/(2sqrt(a_{0)+2)}