What is the tangent of y= 1/(1+x^2),(1,0.5) ?

The tangent of y= 1/(1+x^2),(1,0.5) is y=-1/2 x+1

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tangent of y= 1/(1+x^2),(1,0.5)

tangent of

y = \frac{1}{1 + x ^{2}}, (1, 0.5)

y = - \frac{1}{2} x + 1

Solution steps

Find the slope of

y = \frac{1}{1 + x ^{2}} : \frac{d y}{d x} = - \frac{2 x}{( 1 + x ^{2} ) ^{2}}

Compute the slope of

y = \frac{1}{1 + x ^{2}}

(1, 0.5) : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(1, 0.5) : y = - \frac{1}{2} x + 1

y = - \frac{1}{2} x + 1

y = x^{3} + 6 x - 2, (1, 5)

\frac{d}{d x} (\frac{9}{6 - x})

\int 2 x^{3} + 3 d x

\int_{0}^{1} 0.2 e^{- 0.2 x} d x

What is the tangent of y= 1/(1+x^2),(1,0.5) ?
The tangent of y= 1/(1+x^2),(1,0.5) is y=-1/2 x+1