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

The tangent of \sqrt[3]{x},\at x=8 is y= 1/12 x+4/3

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tangent of \sqrt[3]{x},\at x=8

tangent of

y = \frac{1}{12} x + \frac{4}{3}

Solution steps

Find the tangent point:

(8, 2)

Find the slope of

Compute the slope of

x = 8 : m = \frac{1}{12}

Find the line with slope m=

\frac{1}{12}

and passing through

(8, 2) : y = \frac{1}{12} x + \frac{4}{3}

y = \frac{1}{12} x + \frac{4}{3}

\frac{\partial}{\partial y} (3 x y^{2} e^{4 yz})

x \to 0 + lim (tan^{x} (8 x))

f (x) = x^{2}, (- 1, 1)

\int x cos (2 x^{2} + 7) d x

h \to 0 lim (\frac{( 2. 7 ^{h} - 1 )}{h})

What is the tangent of \sqrt[3]{x},\at x=8 ?
The tangent of \sqrt[3]{x},\at x=8 is y= 1/12 x+4/3