What is the tangent of y=8x^{1/2}+4x^{3/2}+4 ?

The tangent of y=8x^{1/2}+4x^{3/2}+4 is y=(4+6a_{0})/(a_{0)^{1/2}}x+4a_{0}^{1/2}-2a_{0}^{3/2}+4

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tangent of y=8x^{1/2}+4x^{3/2}+4

tangent of

y = 8 x^{\frac{1}{2}} + 4 x^{\frac{3}{2}} + 4

y = \frac{4 + 6 a _{0}}{a _{0}^{\frac{1}{2}}} x + 4 a_{0}^{\frac{1}{2}} - 2 a_{0}^{\frac{3}{2}} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{\frac{1}{2}} + 4 a_{0}^{\frac{3}{2}} + 4)

Find the slope of

y = 8 x^{\frac{1}{2}} + 4 x^{\frac{3}{2}} + 4 : \frac{d y}{d x} = \frac{4 + 6 x}{x ^{\frac{1}{2}}}

Compute the slope of

y = 8 x^{\frac{1}{2}} + 4 x^{\frac{3}{2}} + 4

x = a_{0} : m = \frac{4 + 6 a _{0}}{a _{0}^{\frac{1}{2}}}

Find the line with slope m=

\frac{4 + 6 a _{0}}{a _{0}^{\frac{1}{2}}}

and passing through

(a_{0}, 8 a_{0}^{\frac{1}{2}} + 4 a_{0}^{\frac{3}{2}} + 4) : y = \frac{4 + 6 a _{0}}{a _{0}^{\frac{1}{2}}} x + 4 a_{0}^{\frac{1}{2}} - 2 a_{0}^{\frac{3}{2}} + 4

y = \frac{4 + 6 a _{0}}{a _{0}^{\frac{1}{2}}} x + 4 a_{0}^{\frac{1}{2}} - 2 a_{0}^{\frac{3}{2}} + 4

x \to \infty lim (\frac{x - cos ( x )}{x})

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

y = x^{2} - 8 x - 6

\frac{d y}{d x} = x e^{6 x - 5 y}

\frac{\partial}{\partial y} (e^{2 x})

What is the tangent of y=8x^{1/2}+4x^{3/2}+4 ?
The tangent of y=8x^{1/2}+4x^{3/2}+4 is y=(4+6a_{0})/(a_{0)^{1/2}}x+4a_{0}^{1/2}-2a_{0}^{3/2}+4