What is the tangent of \sqrt[5]{x} ?

The tangent of \sqrt[5]{x} is y= 1/(5a_{0)^{4/5}}x+\sqrt[5]{a_{0}}-(a_{0}^{1/5})/5

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tangent of \sqrt[5]{x}

tangent of

5 x

y = \frac{1}{5 a _{0}^{\frac{4}{5}}} x + 5 a_{0} - \frac{a _{0}^{\frac{1}{5}}}{5}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0})

Find the slope of

y = 5 x : \frac{d y}{d x} = \frac{1}{5 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{1}{5 a _{0}^{\frac{4}{5}}}

Find the line with slope m=

\frac{1}{5 a _{0}^{\frac{4}{5}}}

and passing through

(a_{0}, 5 a_{0}) : y = \frac{1}{5 a _{0}^{\frac{4}{5}}} x + 5 a_{0} - \frac{a _{0}^{\frac{1}{5}}}{5}

y = \frac{1}{5 a _{0}^{\frac{4}{5}}} x + 5 a_{0} - \frac{a _{0}^{\frac{1}{5}}}{5}

f (x) = x^{4} - 17 x^{2} + 16

\int_{0}^{1} x^{2} - x^{3} d x

f (x) = 2 x^{2} + 1, (- 1, 3)

f (x) = \frac{e ^{x}}{9 x ^{2} + 2}

\int \frac{1}{u ^{2}} d u

What is the tangent of \sqrt[5]{x} ?
The tangent of \sqrt[5]{x} is y= 1/(5a_{0)^{4/5}}x+\sqrt[5]{a_{0}}-(a_{0}^{1/5})/5