What is the tangent of f(x)=-8(4x+3)^{-3},\at x=3 ?

The tangent of f(x)=-8(4x+3)^{-3},\at x=3 is y= 32/16875 x-136/16875

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tangent of f(x)=-8(4x+3)^{-3},\at x=3

tangent of

f (x) = - 8 (4 x + 3)^{- 3}, at x = 3

y = \frac{32}{16875} x - \frac{136}{16875}

Solution steps

Find the tangent point:

(3, - \frac{8}{3375})

Find the slope of

f (x) = - 8 (4 x + 3)^{- 3} : \frac{df ( x )}{d x} = \frac{96}{( 4 x + 3 ) ^{4}}

Compute the slope of

f (x) = - 8 (4 x + 3)^{- 3}

x = 3 : m = \frac{32}{16875}

Find the line with slope m=

\frac{32}{16875}

and passing through

(3, - \frac{8}{3375}) : y = \frac{32}{16875} x - \frac{136}{16875}

y = \frac{32}{16875} x - \frac{136}{16875}

\frac{d}{d x} (cot (x) + csc (x))

x \to 1 lim (\frac{1 - x ^{2}}{ln ( x )})

y^{^{'}} + \frac{4 y}{x} = - \frac{3}{x}

\frac{d}{d x} (2 e^{- 1.5 x} cos (3 x))

What is the tangent of f(x)=-8(4x+3)^{-3},\at x=3 ?
The tangent of f(x)=-8(4x+3)^{-3},\at x=3 is y= 32/16875 x-136/16875