What is the tangent of f(x)=(36x)/(x^2+36),\at x=-3 ?

The tangent of f(x)=(36x)/(x^2+36),\at x=-3 is y= 12/25 x-24/25

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tangent of f(x)=(36x)/(x^2+36),\at x=-3

tangent of

f (x) = \frac{36 x}{x ^{2} + 36}, at x = - 3

y = \frac{12}{25} x - \frac{24}{25}

Solution steps

Find the tangent point:

(- 3, - \frac{12}{5})

Find the slope of

f (x) = \frac{36 x}{x ^{2} + 36} : \frac{df ( x )}{d x} = \frac{36 ( - x ^{2} + 36 )}{( x ^{2} + 36 ) ^{2}}

Compute the slope of

f (x) = \frac{36 x}{x ^{2} + 36}

x = - 3 : m = \frac{12}{25}

Find the line with slope m=

\frac{12}{25}

and passing through

(- 3, - \frac{12}{5}) : y = \frac{12}{25} x - \frac{24}{25}

y = \frac{12}{25} x - \frac{24}{25}

\int \frac{x}{( 1 + x ) ^{3}} d x

\frac{\partial}{\partial x} (e^{3 x} cos (4 x))

\frac{d}{d t} (3 + 3 cos (t^{3}) - 4 sin (t^{3}))

x^{- 10}

y^{^{'}} + 5 y = 2 e^{(6 \cdot x)}

What is the tangent of f(x)=(36x)/(x^2+36),\at x=-3 ?
The tangent of f(x)=(36x)/(x^2+36),\at x=-3 is y= 12/25 x-24/25