What is the tangent of f(x)=x^2-x^4,(1,0) ?

The tangent of f(x)=x^2-x^4,(1,0) is y=-2x+2

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tangent of f(x)=x^2-x^4,(1,0)

tangent of

f (x) = x^{2} - x^{4}, (1, 0)

y = - 2 x + 2

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 0) : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 0) : y = - 2 x + 2

y = - 2 x + 2

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

\frac{d x}{d y} - \frac{x}{y} = 2 y

\frac{d}{d x} y = 24 x

x \to 0 lim (2 x sin (\frac{1}{x}))

y^{^{'}} = \frac{e ^{y} + 1}{e ^{y}}

What is the tangent of f(x)=x^2-x^4,(1,0) ?
The tangent of f(x)=x^2-x^4,(1,0) is y=-2x+2