What is the tangent of (x-y-1)^3=x ?

The tangent of (x-y-1)^3=x is y=(-1/(3a_{0)^{2/3}}+1)x-\sqrt[3]{a_{0}}+(a_{0}^{1/3})/3-1

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tangent of (x-y-1)^3=x

tangent of

(x - y - 1)^{3} = x

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

Find the slope of

(x - y - 1)^{3} = x : \frac{d y}{d x} = - \frac{1}{3 x ^{\frac{2}{3}}} + 1

Compute the slope of

(x - y - 1)^{3} = x

x = a_{0} : m = - \frac{1}{3 a _{0}^{\frac{2}{3}}} + 1

Find the line with slope m=

- \frac{1}{3 a _{0}^{\frac{2}{3}}} + 1

and passing through

n = 1 \sum \infty \frac{x ^{n}}{3 ^{n}}

\frac{d}{d x} (\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9})

x \to 1 lim (\frac{x + 1}{x + 2})

f^{^{'}} (x) = x^{4} - 3 x^{3} - 6 x^{2} + 6 x - 1

\int_{1}^{3} 5 x^{3} ln (x) d x

What is the tangent of (x-y-1)^3=x ?
The tangent of (x-y-1)^3=x is y=(-1/(3a_{0)^{2/3}}+1)x-\sqrt[3]{a_{0}}+(a_{0}^{1/3})/3-1