What is the tangent of (ln(x))^4 ?

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

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tangent of (ln(x))^4

tangent of

(ln (x))^{4}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, ln^{4} (a_{0}))

Find the slope of

y = (ln (x))^{4} : \frac{d y}{d x} = \frac{4 ln ^{3} ( x )}{x}

Compute the slope of

y = (ln (x))^{4}

x = a_{0} : m = \frac{4 ln ^{3} ( a _{0} )}{a _{0}}

Find the line with slope m=

\frac{4 ln ^{3} ( a _{0} )}{a _{0}}

and passing through

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

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

\int 2 + e^{2 x} + e^{- 2 x} d x

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

cos (t) \cdot cos (π)

\int sin^{5} (3 x) cos (3 x) d x

f (x) = x^{2} - 5, (- 4, 11)

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