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

The tangent of f(x)=ln(1-x^2+2x^4),\at x=1 is y=3x+ln(2)-3

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

tangent of

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

y = 3 x + ln (2) - 3

Solution steps

Find the tangent point:

(1, ln (2))

Find the slope of

f (x) = ln (1 - x^{2} + 2 x^{4}) : \frac{df ( x )}{d x} = \frac{8 x ^{3} - 2 x}{1 - x ^{2} + 2 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 = 3

Find the line with slope m=

3

and passing through

(1, ln (2)) : y = 3 x + ln (2) - 3

y = 3 x + ln (2) - 3

\int \frac{1}{( 3 x - 2 ) ^{2}} d x

n = 1 \sum \infty \frac{4}{n ( n + 3 )}

\frac{\partial}{\partial y} (ln (x - 2 y))

\int x ln (2) + \frac{3}{2} x d x

x y^{^{'}} + 2 y = 4 sin (x)

What is the tangent of f(x)=ln(1-x^2+2x^4),\at x=1 ?
The tangent of f(x)=ln(1-x^2+2x^4),\at x=1 is y=3x+ln(2)-3