What is the tangent of x^3+x+12x^{1/2} ?

The tangent of x^3+x+12x^{1/2} is y=((6+3a_{0}^{5/2})/(a_{0)^{1/2}}+1)x-2a_{0}^3+6a_{0}^{1/2}

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tangent of x^3+x+12x^{1/2}

tangent of

x^{3} + x + 12 x^{\frac{1}{2}}

y = (\frac{6 + 3 a _{0}^{\frac{5}{2}}}{a _{0}^{\frac{1}{2}}} + 1) x - 2 a_{0}^{3} + 6 a_{0}^{\frac{1}{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + a_{0} + 12 a_{0}^{\frac{1}{2}})

Find the slope of

y = x^{3} + x + 12 x^{\frac{1}{2}} : \frac{d y}{d x} = \frac{6 + 3 x ^{\frac{5}{2}}}{x ^{\frac{1}{2}}} + 1

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{6 + 3 a _{0}^{\frac{5}{2}}}{a _{0}^{\frac{1}{2}}} + 1

Find the line with slope m=

\frac{6 + 3 a _{0}^{\frac{5}{2}}}{a _{0}^{\frac{1}{2}}} + 1

and passing through

(a_{0}, a_{0}^{3} + a_{0} + 12 a_{0}^{\frac{1}{2}}) : y = (\frac{6 + 3 a _{0}^{\frac{5}{2}}}{a _{0}^{\frac{1}{2}}} + 1) x - 2 a_{0}^{3} + 6 a_{0}^{\frac{1}{2}}

y = (\frac{6 + 3 a _{0}^{\frac{5}{2}}}{a _{0}^{\frac{1}{2}}} + 1) x - 2 a_{0}^{3} + 6 a_{0}^{\frac{1}{2}}

\frac{d}{d x} ((5 x + 1)^{3} (6 x^{2} + 3 x + 2)^{5})

\frac{d}{d x} (\frac{sec ^{2} ( x ^{3} )}{2})^{4}

\frac{d}{d x} (4 x (3 x - 2)^{5})

x \to 0 lim (x - 1)

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

What is the tangent of x^3+x+12x^{1/2} ?
The tangent of x^3+x+12x^{1/2} is y=((6+3a_{0}^{5/2})/(a_{0)^{1/2}}+1)x-2a_{0}^3+6a_{0}^{1/2}