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

The tangent of y=2x^2+3/x is y=(4a_{0}-3/(a_{0)^2})x+6/(a_{0)}-2a_{0}^2

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tangent of y=2x^2+3/x

tangent of

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

y = (4 a_{0} - \frac{3}{a _{0}^{2}}) x + \frac{6}{a _{0}} - 2 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 2 x^{2} + \frac{3}{x} : \frac{d y}{d x} = 4 x - \frac{3}{x ^{2}}

Compute the slope of

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

x = a_{0} : m = 4 a_{0} - \frac{3}{a _{0}^{2}}

Find the line with slope m=

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

and passing through

(a_{0}, 2 a_{0}^{2} + \frac{3}{a _{0}}) : y = (4 a_{0} - \frac{3}{a _{0}^{2}}) x + \frac{6}{a _{0}} - 2 a_{0}^{2}

y = (4 a_{0} - \frac{3}{a _{0}^{2}}) x + \frac{6}{a _{0}} - 2 a_{0}^{2}

\frac{d}{d t} (0.1 cos (1 0^{5}) t)

\frac{d y}{d x} + \frac{y}{x} = \frac{1}{x ^{2}}

\int e^{- 4 x + π} d x

\int (x - 5) sin (x) d x

\int (\frac{( x + 1 )}{x}) d x

What is the tangent of y=2x^2+3/x ?
The tangent of y=2x^2+3/x is y=(4a_{0}-3/(a_{0)^2})x+6/(a_{0)}-2a_{0}^2