What is the tangent of f(x)= 3/(1+x^2),(1, 3/2) ?

The tangent of f(x)= 3/(1+x^2),(1, 3/2) is y=-3/2 x+3

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tangent of f(x)= 3/(1+x^2),(1, 3/2)

tangent of

f (x) = \frac{3}{1 + x ^{2}}, (1, \frac{3}{2})

y = - \frac{3}{2} x + 3

Solution steps

Find the slope of

f (x) = \frac{3}{1 + x ^{2}} : \frac{df ( x )}{d x} = - \frac{6 x}{( 1 + x ^{2} ) ^{2}}

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 = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(1, \frac{3}{2}) : y = - \frac{3}{2} x + 3

y = - \frac{3}{2} x + 3

\int \frac{3 x ^{3}}{1 + x ^{4}} d x

\frac{d}{d x} (\frac{5}{4 + x})

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{8}})

f (x) = x^{2} - 12, g (x) = 4 x

\frac{d}{d x} (\frac{cos ( x ) - 1}{x})

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