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

The tangent of (-1)/((2x+1)sqrt(2x+1)) is y= 3/((2a_{0)+1)^{5/2}}x-(5a_{0}+1)/((2a_{0)+1)^{5/2}}

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tangent of (-1)/((2x+1)sqrt(2x+1))

tangent of

\frac{- 1}{( 2 x + 1 ) 2 x + 1}

y = \frac{3}{( 2 a _{0} + 1 ) ^{\frac{5}{2}}} x - \frac{5 a _{0} + 1}{( 2 a _{0} + 1 ) ^{\frac{5}{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 ) 2 a _{0} + 1})

Find the slope of

y = \frac{- 1}{( 2 x + 1 ) 2 x + 1} : \frac{d y}{d x} = \frac{3}{( 2 x + 1 ) ^{\frac{5}{2}}}

Compute the slope of

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

x = a_{0} : m = \frac{3}{( 2 a _{0} + 1 ) ^{\frac{5}{2}}}

Find the line with slope m=

\frac{3}{( 2 a _{0} + 1 ) ^{\frac{5}{2}}}

and passing through

(a_{0}, - \frac{1}{( 2 a _{0} + 1 ) 2 a _{0} + 1}) : y = \frac{3}{( 2 a _{0} + 1 ) ^{\frac{5}{2}}} x - \frac{5 a _{0} + 1}{( 2 a _{0} + 1 ) ^{\frac{5}{2}}}

y = \frac{3}{( 2 a _{0} + 1 ) ^{\frac{5}{2}}} x - \frac{5 a _{0} + 1}{( 2 a _{0} + 1 ) ^{\frac{5}{2}}}

\frac{d}{d x} (1 - 3 x)

\frac{d}{d x} (x^{3} + 3 x^{2} - 9 x + 1)

\frac{d}{d x} (e^{8 - x})

\int \frac{1}{x ^{2} + 6 x + 45} d x

\frac{d}{d x} (\frac{1}{( 5 + x ) ^{2}})

What is the tangent of (-1)/((2x+1)sqrt(2x+1)) ?
The tangent of (-1)/((2x+1)sqrt(2x+1)) is y= 3/((2a_{0)+1)^{5/2}}x-(5a_{0}+1)/((2a_{0)+1)^{5/2}}