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

The tangent of (4x+4)^{1/3},\at x=2 is y= 4/(3*12^{2/3)}x+12^{1/3}-(2*2^{2/3})/(3*3^{2/3)}

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

tangent of

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

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

Solution steps

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = (4 x + 4)^{\frac{1}{3}}

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

Find the line with slope m=

\frac{4}{3 \cdot 1 2 ^{\frac{2}{3}}}

and passing through

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

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

y = \frac{- 6 x}{x ^{2} + 1}, (- 1, 3)

y^{^{''}} + 4 y^{^{'}} - 32 y = 0

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

\frac{d}{d x} (h (g (x) k (x)))

\frac{( s + 1 )}{s ^{2} + s + 1}

What is the tangent of (4x+4)^{1/3},\at x=2 ?
The tangent of (4x+4)^{1/3},\at x=2 is y= 4/(3*12^{2/3)}x+12^{1/3}-(2*2^{2/3})/(3*3^{2/3)}