What is the tangent of (8x+9)^{-2} ?

The tangent of (8x+9)^{-2} is y=-(16)/((8a_{0)+9)^3}x+(24a_{0}+9)/((8a_{0)+9)^3}

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tangent of (8x+9)^{-2}

tangent of

(8 x + 9)^{- 2}

y = - \frac{16}{( 8 a _{0} + 9 ) ^{3}} x + \frac{24 a _{0} + 9}{( 8 a _{0} + 9 ) ^{3}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{( 8 a _{0} + 9 ) ^{2}})

Find the slope of

y = (8 x + 9)^{- 2} : \frac{d y}{d x} = - \frac{16}{( 8 x + 9 ) ^{3}}

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{16}{( 8 a _{0} + 9 ) ^{3}}

Find the line with slope m=

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

and passing through

(a_{0}, \frac{1}{( 8 a _{0} + 9 ) ^{2}}) : y = - \frac{16}{( 8 a _{0} + 9 ) ^{3}} x + \frac{24 a _{0} + 9}{( 8 a _{0} + 9 ) ^{3}}

y = - \frac{16}{( 8 a _{0} + 9 ) ^{3}} x + \frac{24 a _{0} + 9}{( 8 a _{0} + 9 ) ^{3}}

What is the tangent of (8x+9)^{-2} ?
The tangent of (8x+9)^{-2} is y=-(16)/((8a_{0)+9)^3}x+(24a_{0}+9)/((8a_{0)+9)^3}