What is the tangent of x^2+4/x-10 ?

The tangent of x^2+4/x-10 is y=(2a_{0}-4/(a_{0)^2})x+8/(a_{0)}-a_{0}^2-10

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tangent of x^2+4/x-10

tangent of

x^{2} + \frac{4}{x} - 10

y = (2 a_{0} - \frac{4}{a _{0}^{2}}) x + \frac{8}{a _{0}} - a_{0}^{2} - 10

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + \frac{4}{a _{0}} - 10)

Find the slope of

y = x^{2} + \frac{4}{x} - 10 : \frac{d y}{d x} = 2 x - \frac{4}{x ^{2}}

Compute the slope of

y = x^{2} + \frac{4}{x} - 10

x = a_{0} : m = 2 a_{0} - \frac{4}{a _{0}^{2}}

Find the line with slope m=

2 a_{0} - \frac{4}{a _{0}^{2}}

and passing through

(a_{0}, a_{0}^{2} + \frac{4}{a _{0}} - 10) : y = (2 a_{0} - \frac{4}{a _{0}^{2}}) x + \frac{8}{a _{0}} - a_{0}^{2} - 10

y = (2 a_{0} - \frac{4}{a _{0}^{2}}) x + \frac{8}{a _{0}} - a_{0}^{2} - 10

3 x - 4 x^{2}

\int (x^{1.8} + 9 x^{3.5}) d x

(4 y + x^{2} y) d y - (2 x + x y^{2}) d x = 0

y^{^{''}} - y^{^{'}} - 2 y = 8 e^{- t}

\frac{d}{d x} ((- 2)^{x})

What is the tangent of x^2+4/x-10 ?
The tangent of x^2+4/x-10 is y=(2a_{0}-4/(a_{0)^2})x+8/(a_{0)}-a_{0}^2-10