What is the tangent of f(x)=2\sqrt[3]{x},\at x=8 ?

The tangent of f(x)=2\sqrt[3]{x},\at x=8 is y= 1/6 x+8/3

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tangent of f(x)=2\sqrt[3]{x},\at x=8

tangent of

y = \frac{1}{6} x + \frac{8}{3}

Solution steps

Find the tangent point:

(8, 4)

Find the slope of

Compute the slope of

x = 8 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(8, 4) : y = \frac{1}{6} x + \frac{8}{3}

y = \frac{1}{6} x + \frac{8}{3}

\frac{d x}{d t} = \frac{( e ^{t} - e ^{- t} )}{3 + x}

y^{^{'}} + \frac{1}{x} y = 10 x^{2}

\frac{d}{d x} (1 - sin (x))

\frac{d}{d x} (6 x - 5)

\frac{d}{d x} (\frac{e ^{2 x} + e ^{- 2 x}}{x})

What is the tangent of f(x)=2\sqrt[3]{x},\at x=8 ?
The tangent of f(x)=2\sqrt[3]{x},\at x=8 is y= 1/6 x+8/3