What is the tangent of f(x)=(4x-4)^{1/3},\at x=2 ?

The tangent of f(x)=(4x-4)^{1/3},\at x=2 is y=(4^{1/3})/3 x+4^{1/3}-(2*4^{1/3})/3

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tangent of f(x)=(4x-4)^{1/3},\at x=2

tangent of

f (x) = (4 x - 4)^{\frac{1}{3}}, at x = 2

y = \frac{4 ^{\frac{1}{3}}}{3} x + 4^{\frac{1}{3}} - \frac{2 \cdot 4 ^{\frac{1}{3}}}{3}

Solution steps

Find the tangent point:

(2, 4^{\frac{1}{3}})

Find the slope of

f (x) = (4 x - 4)^{\frac{1}{3}} : \frac{df ( x )}{d x} = \frac{4}{3 ( 4 x - 4 ) ^{\frac{2}{3}}}

Compute the slope of

f (x) = (4 x - 4)^{\frac{1}{3}}

x = 2 : m = \frac{4 ^{\frac{1}{3}}}{3}

Find the line with slope m=

\frac{4 ^{\frac{1}{3}}}{3}

and passing through

(2, 4^{\frac{1}{3}}) : y = \frac{4 ^{\frac{1}{3}}}{3} x + 4^{\frac{1}{3}} - \frac{2 \cdot 4 ^{\frac{1}{3}}}{3}

y = \frac{4 ^{\frac{1}{3}}}{3} x + 4^{\frac{1}{3}} - \frac{2 \cdot 4 ^{\frac{1}{3}}}{3}

f (x) = arctan (e^{x})

x^{^{''}} + x^{^{'}} + x = 0

t y^{^{'}} - 2 y = 2 t^{4}

\frac{d}{d x} (\frac{x - 7}{- 5 x - 10})

\frac{\partial}{\partial x} (x y + y z^{2} + x z^{3})

What is the tangent of f(x)=(4x-4)^{1/3},\at x=2 ?
The tangent of f(x)=(4x-4)^{1/3},\at x=2 is y=(4^{1/3})/3 x+4^{1/3}-(2*4^{1/3})/3