What is the tangent of-(16)/((8x+9)^3) ?

The tangent of-(16)/((8x+9)^3) is y=(384)/((8a_{0)+9)^4}x+(-512a_{0}-144)/((8a_{0)+9)^4}

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tangent of-(16)/((8x+9)^3)

tangent of

- \frac{16}{( 8 x + 9 ) ^{3}}

y = \frac{384}{( 8 a _{0} + 9 ) ^{4}} x + \frac{- 512 a _{0} - 144}{( 8 a _{0} + 9 ) ^{4}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{16}{( 8 a _{0} + 9 ) ^{3}})

Find the slope of

y = - \frac{16}{( 8 x + 9 ) ^{3}} : \frac{d y}{d x} = \frac{384}{( 8 x + 9 ) ^{4}}

Compute the slope of

y = - \frac{16}{( 8 x + 9 ) ^{3}}

x = a_{0} : m = \frac{384}{( 8 a _{0} + 9 ) ^{4}}

Find the line with slope m=

\frac{384}{( 8 a _{0} + 9 ) ^{4}}

and passing through

(a_{0}, - \frac{16}{( 8 a _{0} + 9 ) ^{3}}) : y = \frac{384}{( 8 a _{0} + 9 ) ^{4}} x + \frac{- 512 a _{0} - 144}{( 8 a _{0} + 9 ) ^{4}}

y = \frac{384}{( 8 a _{0} + 9 ) ^{4}} x + \frac{- 512 a _{0} - 144}{( 8 a _{0} + 9 ) ^{4}}

\frac{d y}{d x} = \frac{y ^{2} + y}{x ^{2} + x}

\int_{0}^{1} x e^{- 2 x} d x

y = 4 x^{2} - 1, y = 0, x = \frac{- 1}{2}, x = 1

\int \frac{x + 25}{x ^{2} + 24 x + 148} d x

y^{^{'}} = \frac{( cos ( 18 x ) )}{( e ^{(18 \cdot y)} )}

What is the tangent of-(16)/((8x+9)^3) ?
The tangent of-(16)/((8x+9)^3) is y=(384)/((8a_{0)+9)^4}x+(-512a_{0}-144)/((8a_{0)+9)^4}