What is the tangent of f(x)=ln(6-9x^2+8x^4),\at x=1 ?

The tangent of f(x)=ln(6-9x^2+8x^4),\at x=1 is y= 14/5 x+ln(5)-14/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=ln(6-9x^2+8x^4),\at x=1

tangent of

f (x) = ln (6 - 9 x^{2} + 8 x^{4}), at x = 1

y = \frac{14}{5} x + ln (5) - \frac{14}{5}

Solution steps

Find the tangent point:

(1, ln (5))

Find the slope of

f (x) = ln (6 - 9 x^{2} + 8 x^{4}) : \frac{df ( x )}{d x} = \frac{32 x ^{3} - 18 x}{6 - 9 x ^{2} + 8 x ^{4}}

Compute the slope of

f (x) = ln (6 - 9 x^{2} + 8 x^{4})

x = 1 : m = \frac{14}{5}

Find the line with slope m=

\frac{14}{5}

and passing through

(1, ln (5)) : y = \frac{14}{5} x + ln (5) - \frac{14}{5}

y = \frac{14}{5} x + ln (5) - \frac{14}{5}

\int \frac{r ^{3}}{4 - r ^{2}} d r

y = \frac{9}{x - 5} - 1

\int \frac{x}{x ^{4} + 4} d x

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

\frac{d}{d x} (tan (\frac{x}{5}))

What is the tangent of f(x)=ln(6-9x^2+8x^4),\at x=1 ?
The tangent of f(x)=ln(6-9x^2+8x^4),\at x=1 is y= 14/5 x+ln(5)-14/5