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}}}

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{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} (e^{8 - x})

d er i v a t i v e x^{2} + 1 + x^{2}

\int (2 t - 1)^{2} d t

\frac{d}{d x} (3 \cdot cos (x))

\frac{d}{d x} (\frac{3 x ^{2}}{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}}