What is the tangent of x^3+x^2+6/x ?

The tangent of x^3+x^2+6/x is y=(3a_{0}^2+2a_{0}-6/(a_{0)^2})x-2a_{0}^3-a_{0}^2+(12)/(a_{0)}

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tangent of x^3+x^2+6/x

tangent of

x^{3} + x^{2} + \frac{6}{x}

y = (3 a_{0}^{2} + 2 a_{0} - \frac{6}{a _{0}^{2}}) x - 2 a_{0}^{3} - a_{0}^{2} + \frac{12}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + a_{0}^{2} + \frac{6}{a _{0}})

Find the slope of

y = x^{3} + x^{2} + \frac{6}{x} : \frac{d y}{d x} = 3 x^{2} + 2 x - \frac{6}{x ^{2}}

Compute the slope of

y = x^{3} + x^{2} + \frac{6}{x}

x = a_{0} : m = 3 a_{0}^{2} + 2 a_{0} - \frac{6}{a _{0}^{2}}

Find the line with slope m=

3 a_{0}^{2} + 2 a_{0} - \frac{6}{a _{0}^{2}}

and passing through

(a_{0}, a_{0}^{3} + a_{0}^{2} + \frac{6}{a _{0}}) : y = (3 a_{0}^{2} + 2 a_{0} - \frac{6}{a _{0}^{2}}) x - 2 a_{0}^{3} - a_{0}^{2} + \frac{12}{a _{0}}

y = (3 a_{0}^{2} + 2 a_{0} - \frac{6}{a _{0}^{2}}) x - 2 a_{0}^{3} - a_{0}^{2} + \frac{12}{a _{0}}

\int \frac{8 x}{x ^{2} - 3} d x

\int \frac{1}{sin ^{2} ( 2 x ) cos ^{2} ( 2 x )} d x

\int sec^{3} (x) - sec (x) d x

\int_{0}^{1} (\frac{9 t}{1 + t ^{2}}) d t

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

What is the tangent of x^3+x^2+6/x ?
The tangent of x^3+x^2+6/x is y=(3a_{0}^2+2a_{0}-6/(a_{0)^2})x-2a_{0}^3-a_{0}^2+(12)/(a_{0)}