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

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

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

tangent of

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

y = \frac{6}{5} x + ln (5) - \frac{6}{5}

Solution steps

Find the tangent point:

(1, ln (5))

Find the slope of

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

Compute the slope of

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

x = 1 : m = \frac{6}{5}

Find the line with slope m=

\frac{6}{5}

and passing through

(1, ln (5)) : y = \frac{6}{5} x + ln (5) - \frac{6}{5}

y = \frac{6}{5} x + ln (5) - \frac{6}{5}

f (t) = \frac{9}{4} t^{2} \cdot e^{2 t} \cdot e^{3 t}

\int 128 cos^{4} (8 x) d x

\int y^{2} y d y

\frac{d}{d x} (\frac{8 x ^{2} + 2 x + 6}{x})

\int_{2}^{5} x^{3} d x

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