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

The tangent of f(x)= 9/(5+x^2),\at x=4 is y=-8/49 x+53/49

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tangent of f(x)= 9/(5+x^2),\at x=4

tangent of

f (x) = \frac{9}{5 + x ^{2}}, at x = 4

y = - \frac{8}{49} x + \frac{53}{49}

Solution steps

Find the tangent point:

(4, \frac{3}{7})

Find the slope of

f (x) = \frac{9}{5 + x ^{2}} : \frac{df ( x )}{d x} = - \frac{18 x}{( 5 + x ^{2} ) ^{2}}

Compute the slope of

f (x) = \frac{9}{5 + x ^{2}}

x = 4 : m = - \frac{8}{49}

Find the line with slope m=

- \frac{8}{49}

and passing through

(4, \frac{3}{7}) : y = - \frac{8}{49} x + \frac{53}{49}

y = - \frac{8}{49} x + \frac{53}{49}

f (x) = x^{2} + 3, at x = 6

\int \frac{cosh ( x )}{x} d x

\frac{4 - 4 s}{4 s ^{2} + 4 s + 17}

f (x) = sec (x) + tan (x)

\int 8 x - 4 d x

What is the tangent of f(x)= 9/(5+x^2),\at x=4 ?
The tangent of f(x)= 9/(5+x^2),\at x=4 is y=-8/49 x+53/49