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

The tangent of f(x)= 1/(1+5x^2),\at x=-1 is y= 5/18 x+4/9

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tangent of f(x)= 1/(1+5x^2),\at x=-1

tangent of

f (x) = \frac{1}{1 + 5 x ^{2}}, at x = - 1

y = \frac{5}{18} x + \frac{4}{9}

Solution steps

Find the tangent point:

(- 1, \frac{1}{6})

Find the slope of

f (x) = \frac{1}{1 + 5 x ^{2}} : \frac{df ( x )}{d x} = - \frac{10 x}{( 1 + 5 x ^{2} ) ^{2}}

Compute the slope of

f (x) = \frac{1}{1 + 5 x ^{2}}

x = - 1 : m = \frac{5}{18}

Find the line with slope m=

\frac{5}{18}

and passing through

(- 1, \frac{1}{6}) : y = \frac{5}{18} x + \frac{4}{9}

y = \frac{5}{18} x + \frac{4}{9}

\int x^{2} (sin (π x)) d x

\frac{d}{d t} (7 t)

\frac{\partial}{\partial x} (\frac{x}{y} - sin (y))

x y^{^{''}} + 2 y^{^{'}} = 12 x^{2}

\int_{- 1}^{1} \frac{4 x}{( x + 3 ) ^{2}} d x

What is the tangent of f(x)= 1/(1+5x^2),\at x=-1 ?
The tangent of f(x)= 1/(1+5x^2),\at x=-1 is y= 5/18 x+4/9