What is the tangent of y= 9/(x^2+2),(1,3) ?

The tangent of y= 9/(x^2+2),(1,3) is y=-2x+5

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tangent of y= 9/(x^2+2),(1,3)

tangent of

y = \frac{9}{x ^{2} + 2}, (1, 3)

y = - 2 x + 5

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{9}{x ^{2} + 2}

(1, 3) : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 3) : y = - 2 x + 5

y = - 2 x + 5

\frac{d}{d y} (ln (\frac{( 5 y + 1 ) ^{2}}{y ^{2} + 1}))

ln (4 x^{3} + x)

\int 5 5 x d x

\frac{\partial}{\partial x} (sin (x) - 3 cos (y))

\frac{\partial}{\partial x} (\frac{sin ( t )}{t})

What is the tangent of y= 9/(x^2+2),(1,3) ?
The tangent of y= 9/(x^2+2),(1,3) is y=-2x+5