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

4 x, 4 x

\int - 0.5 x d x

x \to 0 lim (\frac{e ^{6 x} - 1}{x})

\frac{\partial}{\partial x} (\frac{( a R ( x ) + b + c R ( x ) + d e + f ( x ) + g ( x ) e )}{x})

y ln (x) y^{^{'}} = (\frac{( y + 1 )}{x ^{(\frac{1}{2})}})^{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