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

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

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

tangent of

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

y = \frac{5}{2} x + 5

Solution steps

Find the tangent point:

(- 1, \frac{5}{2})

Find the slope of

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

Compute the slope of

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

x = - 1 : m = \frac{5}{2}

Find the line with slope m=

\frac{5}{2}

and passing through

(- 1, \frac{5}{2}) : y = \frac{5}{2} x + 5

y = \frac{5}{2} x + 5

y^{^{'}} = 2 e^{(6 x - y)}

\int (x^{2} + x) e^{2 x} d x

\frac{\partial}{\partial z} (2 yz)

f (x) = x - 4 tan (x), at x = \frac{3 π}{4}

\int (- 6 sin (2 x) + 3 cos (3 x)) d x

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