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

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

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

tangent of

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

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

Solution steps

Find the slope of

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

Compute the slope of

(x^{2} + 4) y = 8

(- 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

\frac{( s + 2 )}{s ^{2} + s + 1}

\frac{d}{d x} (\frac{d y}{d x}) + 9 y = 0

\int e^{x} (x + 1) cos (x e^{x}) d x

y = (3 x - 4)^{5}

\frac{d}{d x} (ln (2 x + 1) - x^{2})

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