What is the tangent of f(x)=2\sqrt[3]{x},\at x=64 ?

The tangent of f(x)=2\sqrt[3]{x},\at x=64 is y= 1/24 x+16/3

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tangent of f(x)=2\sqrt[3]{x},\at x=64

tangent of

y = \frac{1}{24} x + \frac{16}{3}

Solution steps

Find the tangent point:

(64, 8)

Find the slope of

Compute the slope of

x = 64 : m = \frac{1}{24}

Find the line with slope m=

\frac{1}{24}

and passing through

(64, 8) : y = \frac{1}{24} x + \frac{16}{3}

y = \frac{1}{24} x + \frac{16}{3}

\frac{d}{d x} (x^{5} - 20 x^{3} + 64 x)

\frac{d}{d x} (\frac{10 x ^{3}}{x ^{2} - 1})

\int \frac{( x ^{2} - 16 )}{x} d x

\frac{d}{d x} (arctan (ln (x)))

x \to 0 lim ((1 + x)^{(\frac{1}{x})})

What is the tangent of f(x)=2\sqrt[3]{x},\at x=64 ?
The tangent of f(x)=2\sqrt[3]{x},\at x=64 is y= 1/24 x+16/3