What is the tangent of f(x)= 1/(2sqrt(x+1)) ?

The tangent of f(x)= 1/(2sqrt(x+1)) is y=-1/(4(a_{0)+1)^{3/2}}x+(3a_{0}+2)/(4(a_{0)+1)^{3/2}}

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tangent of f(x)= 1/(2sqrt(x+1))

tangent of

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

y = - \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 2}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{2 a _{0} + 1})

Find the slope of

f (x) = \frac{1}{2 x + 1} : \frac{df ( x )}{d x} = - \frac{1}{4 ( x + 1 ) ^{\frac{3}{2}}}

Compute the slope of

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

x = a_{0} : m = - \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{2 a _{0} + 1}) : y = - \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 2}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

y = - \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 2}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

\frac{d}{d t} (5 cos (2 t))

x \to 0 lim (\frac{tan ( x )}{3 x})

f (x) = - 3 x^{2} - 5 x + 1

\int cos^{4} (t) d t

y d y = 4 x (y^{2} + 1)^{\frac{1}{2}} d x

What is the tangent of f(x)= 1/(2sqrt(x+1)) ?
The tangent of f(x)= 1/(2sqrt(x+1)) is y=-1/(4(a_{0)+1)^{3/2}}x+(3a_{0}+2)/(4(a_{0)+1)^{3/2}}