What is the tangent of f(x)=(9x-1)^{-1/3},\at x=1 ?

The tangent of f(x)=(9x-1)^{-1/3},\at x=1 is y=-3/16 x+11/16

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tangent of f(x)=(9x-1)^{-1/3},\at x=1

tangent of

f (x) = (9 x - 1)^{- \frac{1}{3}}, at x = 1

y = - \frac{3}{16} x + \frac{11}{16}

Solution steps

Find the tangent point:

(1, \frac{1}{2})

Find the slope of

f (x) = (9 x - 1)^{- \frac{1}{3}} : \frac{df ( x )}{d x} = - \frac{3}{( 9 x - 1 ) ^{\frac{4}{3}}}

Compute the slope of

f (x) = (9 x - 1)^{- \frac{1}{3}}

x = 1 : m = - \frac{3}{16}

Find the line with slope m=

- \frac{3}{16}

and passing through

(1, \frac{1}{2}) : y = - \frac{3}{16} x + \frac{11}{16}

y = - \frac{3}{16} x + \frac{11}{16}

t, \frac{d y}{d t} = y^{3} - y^{2} - 12 y

f (x) = \frac{4 e ^{x} + 5 e ^{- x}}{6}

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 4})

\frac{d y}{d x} = \frac{x ^{3}}{y + 1}

x y^{^{'}} - y = x^{2} + y^{2}

What is the tangent of f(x)=(9x-1)^{-1/3},\at x=1 ?
The tangent of f(x)=(9x-1)^{-1/3},\at x=1 is y=-3/16 x+11/16