What is the tangent of (64)/(x^2+16),\at x=3 ?

The tangent of (64)/(x^2+16),\at x=3 is y=-384/625 x+2752/625

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tangent of (64)/(x^2+16),\at x=3

tangent of

\frac{64}{x ^{2} + 16}, at x = 3

y = - \frac{384}{625} x + \frac{2752}{625}

Solution steps

Find the tangent point:

(3, \frac{64}{25})

Find the slope of

y = \frac{64}{x ^{2} + 16} : \frac{d y}{d x} = - \frac{128 x}{( x ^{2} + 16 ) ^{2}}

Compute the slope of

y = \frac{64}{x ^{2} + 16}

x = 3 : m = - \frac{384}{625}

Find the line with slope m=

- \frac{384}{625}

and passing through

(3, \frac{64}{25}) : y = - \frac{384}{625} x + \frac{2752}{625}

y = - \frac{384}{625} x + \frac{2752}{625}

\int_{2}^{3} (x^{2} - 4) d x

f (x) = \frac{6 x}{2 + x ^{2}}, (2, 2)

\int (3 x + 2) d x

y = 19 cos (\frac{x}{b} - 2.5) + 19

\frac{\partial}{\partial x} (sin^{2} (2 x + 3 y))

What is the tangent of (64)/(x^2+16),\at x=3 ?
The tangent of (64)/(x^2+16),\at x=3 is y=-384/625 x+2752/625