What is the tangent of f(x)=ln(6-x^2+2x^4),\at x=1 ?

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

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tangent of f(x)=ln(6-x^2+2x^4),\at x=1

tangent of

f (x) = ln (6 - x^{2} + 2 x^{4}), at x = 1

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

Solution steps

Find the tangent point:

(1, ln (7))

Find the slope of

f (x) = ln (6 - x^{2} + 2 x^{4}) : \frac{df ( x )}{d x} = \frac{8 x ^{3} - 2 x}{6 - x ^{2} + 2 x ^{4}}

Compute the slope of

f (x) = ln (6 - x^{2} + 2 x^{4})

x = 1 : m = \frac{6}{7}

Find the line with slope m=

\frac{6}{7}

and passing through

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

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

\frac{d y}{d x} + (\frac{3}{x}) y = - x + 4

\frac{\partial}{\partial x} (ln (\frac{1}{2 + 3 e ^{x}}))

x \to 1 lim (\frac{3 x - 2}{x - 1})

\frac{1}{3} tan^{3} (x) - tan (x) + x

x \to - 8 lim (\frac{x + 2}{x + 8})

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