What is the tangent of f(x)=4\sqrt[3]{x},\at x=1 ?

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

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tangent of f(x)=4\sqrt[3]{x},\at x=1

tangent of

f (x) = 4 3 x, at x = 1

y = \frac{4}{3} x + \frac{8}{3}

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = 4 3 x : \frac{df ( x )}{d x} = \frac{4}{3 x ^{\frac{2}{3}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{4}{3}

Find the line with slope m=

\frac{4}{3}

and passing through

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

y = \frac{4}{3} x + \frac{8}{3}

\int_{3}^{3} (x + 1) d x

d er i v a t i v e \frac{7}{5 x}

\frac{d}{d x} (x^{2} - 5 x + \frac{1}{x ^{\frac{3}{2}}})

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

\int - 4 x (- 2 x^{2} + 2)^{5} d x

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