What is the tangent of y= 8/(x^2+4),(2,1) ?

The tangent of y= 8/(x^2+4),(2,1) is y=-1/2 x+2

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tangent of y= 8/(x^2+4),(2,1)

tangent of

y = \frac{8}{x ^{2} + 4}, (2, 1)

y = - \frac{1}{2} x + 2

Solution steps

Find the slope of

y = \frac{8}{x ^{2} + 4} : \frac{d y}{d x} = - \frac{16 x}{( x ^{2} + 4 ) ^{2}}

Compute the slope of

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

(2, 1) : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(2, 1) : y = - \frac{1}{2} x + 2

y = - \frac{1}{2} x + 2

\int_{1}^{2} \frac{1}{8} + \frac{3}{8} x d x

\frac{d}{d x} ((3 x^{4} - x + 1) (- x^{5} + 4))

y^{^{'}} + 2 y = 1, y (0) = \frac{5}{2}

\int x (2 x + 3)^{2} d x

\frac{d y}{d x} = \frac{y cos ( x )}{1 - 2 y}

What is the tangent of y= 8/(x^2+4),(2,1) ?
The tangent of y= 8/(x^2+4),(2,1) is y=-1/2 x+2