What is the tangent of y=ln(7-9x^2+8x^4),\at x=1 ?

The tangent of y=ln(7-9x^2+8x^4),\at x=1 is y= 7/3 x+ln(6)-7/3

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tangent of y=ln(7-9x^2+8x^4),\at x=1

tangent of

y = ln (7 - 9 x^{2} + 8 x^{4}), at x = 1

y = \frac{7}{3} x + ln (6) - \frac{7}{3}

Solution steps

Find the tangent point:

(1, ln (6))

Find the slope of

y = ln (7 - 9 x^{2} + 8 x^{4}) : \frac{d y}{d x} = \frac{32 x ^{3} - 18 x}{7 - 9 x ^{2} + 8 x ^{4}}

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{7}{3}

Find the line with slope m=

\frac{7}{3}

and passing through

(1, ln (6)) : y = \frac{7}{3} x + ln (6) - \frac{7}{3}

y = \frac{7}{3} x + ln (6) - \frac{7}{3}

What is the tangent of y=ln(7-9x^2+8x^4),\at x=1 ?
The tangent of y=ln(7-9x^2+8x^4),\at x=1 is y= 7/3 x+ln(6)-7/3