What is the tangent of y=(27)/(x^2+9),\at x=2 ?

The tangent of y=(27)/(x^2+9),\at x=2 is y=-108/169 x+567/169

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tangent of y=(27)/(x^2+9),\at x=2

tangent of

y = \frac{27}{x ^{2} + 9}, at x = 2

y = - \frac{108}{169} x + \frac{567}{169}

Solution steps

Find the tangent point:

(2, \frac{27}{13})

Find the slope of

y = \frac{27}{x ^{2} + 9} : \frac{d y}{d x} = - \frac{54 x}{( x ^{2} + 9 ) ^{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{108}{169}

Find the line with slope m=

- \frac{108}{169}

and passing through

(2, \frac{27}{13}) : y = - \frac{108}{169} x + \frac{567}{169}

y = - \frac{108}{169} x + \frac{567}{169}

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\int_{0}^{\frac{π}{2}} 17 sin^{2} (2 x) d x

x^{4} + 2 x, x = 1

y = \frac{4 x ^{2} + 8 x + 8}{x}

\frac{\partial}{\partial y} (ln (x - 4 y))

What is the tangent of y=(27)/(x^2+9),\at x=2 ?
The tangent of y=(27)/(x^2+9),\at x=2 is y=-108/169 x+567/169