What is the tangent of ln(x^4+3) ?

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

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tangent of ln(x^4+3)

tangent of

ln (x^{4} + 3)

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, ln (a_{0}^{4} + 3))

Find the slope of

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

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{4 a _{0}^{3}}{a _{0}^{4} + 3}

Find the line with slope m=

\frac{4 a _{0}^{3}}{a _{0}^{4} + 3}

and passing through

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

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

\int_{0}^{1} 2 π (x) (e^{- x^{2}}) d x

(3 - x)^{2}

e^{x} - cot (x)

\frac{1}{( s ( s ^{2} + s + 1 ) )}

\int \frac{1}{4 e ^{- 2 x} + e ^{2 x}} d x

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