What is the tangent of ln(x^2-8x+1)(8) ?

The tangent of ln(x^2-8x+1)(8) is y=(8(2a-8))/(a^2-8a+1)x+ln(a^2-8a+1)*8-(8a(2a-8))/(a^2-8a+1)

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tangent of ln(x^2-8x+1)(8)

tangent of

ln (x^{2} - 8 x + 1) (8)

y = \frac{8 ( 2 a - 8 )}{a ^{2} - 8 a + 1} x + ln (a^{2} - 8 a + 1) \cdot 8 - \frac{8 a ( 2 a - 8 )}{a ^{2} - 8 a + 1}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, ln (a^{2} - 8 a + 1) \cdot 8)

Find the slope of

y = ln (x^{2} - 8 x + 1) 8 : \frac{d y}{d x} = \frac{8 ( 2 x - 8 )}{x ^{2} - 8 x + 1}

Compute the slope of

y = ln (x^{2} - 8 x + 1) 8

x = a : m = \frac{8 ( 2 a - 8 )}{a ^{2} - 8 a + 1}

Find the line with slope m=

\frac{8 ( 2 a - 8 )}{a ^{2} - 8 a + 1}

and passing through

(a, ln (a^{2} - 8 a + 1) 8) : y = \frac{8 ( 2 a - 8 )}{a ^{2} - 8 a + 1} x + ln (a^{2} - 8 a + 1) \cdot 8 - \frac{8 a ( 2 a - 8 )}{a ^{2} - 8 a + 1}

y = \frac{8 ( 2 a - 8 )}{a ^{2} - 8 a + 1} x + ln (a^{2} - 8 a + 1) \cdot 8 - \frac{8 a ( 2 a - 8 )}{a ^{2} - 8 a + 1}

\frac{d}{d x} (e^{x} - 2 sin (x))

\int (\frac{cot ^{2} ( x )}{cos ^{2} ( x )}) d x

\frac{( \frac{10}{6} )}{s ( s ^{2} + \frac{1}{2} )}

\frac{5}{6 x ^{5}}

\int (- 3 x^{4} - 6 x^{2} + 8) d x

What is the tangent of ln(x^2-8x+1)(8) ?
The tangent of ln(x^2-8x+1)(8) is y=(8(2a-8))/(a^2-8a+1)x+ln(a^2-8a+1)*8-(8a(2a-8))/(a^2-8a+1)