What is the tangent of f(x)=(6x-2)^{1/3},\at x=1 ?

The tangent of f(x)=(6x-2)^{1/3},\at x=1 is y= 2/(4^{2/3)}x+4^{1/3}-1/(2^{1/3)}

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tangent of f(x)=(6x-2)^{1/3},\at x=1

tangent of

f (x) = (6 x - 2)^{\frac{1}{3}}, at x = 1

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

Solution steps

Find the tangent point:

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

Find the slope of

f (x) = (6 x - 2)^{\frac{1}{3}} : \frac{df ( x )}{d x} = \frac{2}{( 6 x - 2 ) ^{\frac{2}{3}}}

Compute the slope of

f (x) = (6 x - 2)^{\frac{1}{3}}

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

Find the line with slope m=

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

and passing through

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

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

y^{^{'}} - t^{2} y = 0

\int \frac{1}{x ( x - 4 ) ^{2}} d x

\frac{d}{d x} (4 sin (x) cos (x) + 4 sin (x))

\int \frac{1}{( a + b y ) ^{3}} d y

\int \frac{y}{4 - y ^{2}} d y

What is the tangent of f(x)=(6x-2)^{1/3},\at x=1 ?
The tangent of f(x)=(6x-2)^{1/3},\at x=1 is y= 2/(4^{2/3)}x+4^{1/3}-1/(2^{1/3)}