What is the tangent of y=\sqrt[5]{2x^3+8x},(2,2) ?

The tangent of y=\sqrt[5]{2x^3+8x},(2,2) is y= 2/5 x+6/5

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tangent of y=\sqrt[5]{2x^3+8x},(2,2)

tangent of

y = 5 2 x^{3} + 8 x, (2, 2)

y = \frac{2}{5} x + \frac{6}{5}

Solution steps

Find the slope of

y = 5 2 x^{3} + 8 x : \frac{d y}{d x} = \frac{6 x ^{2} + 8}{5 ( 2 x ^{3} + 8 x ) ^{\frac{4}{5}}}

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{2}{5}

Find the line with slope m=

\frac{2}{5}

and passing through

(2, 2) : y = \frac{2}{5} x + \frac{6}{5}

y = \frac{2}{5} x + \frac{6}{5}

x \to 0 lim (\frac{sin ( 0 )}{0})

x^{3} - 3 x^{2} - 9 x + 27, x + 3

x \to 2 lim (x^{2} + 2 x + 3)

y = 4 x + 12, y = x^{2}

\int \frac{1}{( x ^{2} + 6 x + 10 ) ^{2}} d x

What is the tangent of y=\sqrt[5]{2x^3+8x},(2,2) ?
The tangent of y=\sqrt[5]{2x^3+8x},(2,2) is y= 2/5 x+6/5