What is the tangent of y=\sqrt[3]{x},(1,1) ?

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

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tangent of y=\sqrt[3]{x},(1,1)

tangent of

y = \frac{1}{3} x + \frac{2}{3}

Solution steps

Find the slope of

Compute the slope of

(1, 1) : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

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

y = \frac{1}{3} x + \frac{2}{3}

\int w d w

\frac{d}{d x} (6 x^{2} + 4)

\frac{\partial}{\partial x} (ln (x - 8 y))

\int \frac{1}{8 x + 1} - \frac{1}{e ^{x} ( 8 + e ^{- x} ) ^{2}} d x

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