What is the tangent of ln(x^2) ?

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

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tangent of ln(x^2)

tangent of

ln (x^{2})

y = \frac{2}{a _{0}} x + 2 ln (a_{0}) - 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 ln (a_{0}))

Find the slope of

y = ln (x^{2}) : \frac{d y}{d x} = \frac{2}{x}

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{2}{a _{0}}

Find the line with slope m=

\frac{2}{a _{0}}

and passing through

(a_{0}, 2 ln (a_{0})) : y = \frac{2}{a _{0}} x + 2 ln (a_{0}) - 2

y = \frac{2}{a _{0}} x + 2 ln (a_{0}) - 2

\frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}}

x \frac{d y}{d x} = 4 y

y = x^{3}, x = - 1, x = 2, y = 0

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

x \to 2 lim (2 x (x^{2} - 3))

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