What is the tangent of (6x)/(x+8),\at x=1 ?

The tangent of (6x)/(x+8),\at x=1 is y= 16/27 x+2/27

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tangent of (6x)/(x+8),\at x=1

tangent of

\frac{6 x}{x + 8}, at x = 1

y = \frac{16}{27} x + \frac{2}{27}

Solution steps

Find the tangent point:

(1, \frac{2}{3})

Find the slope of

y = \frac{6 x}{x + 8} : \frac{d y}{d x} = \frac{48}{( x + 8 ) ^{2}}

Compute the slope of

y = \frac{6 x}{x + 8}

x = 1 : m = \frac{16}{27}

Find the line with slope m=

\frac{16}{27}

and passing through

(1, \frac{2}{3}) : y = \frac{16}{27} x + \frac{2}{27}

y = \frac{16}{27} x + \frac{2}{27}

\int (5 sec^{2} (x)) d x

x \to 6 - lim (c x^{2} + 5 x)

\frac{\partial}{\partial x} (2 - 2 x^{2} + y^{2})

(5.2) (6.4)

x \to - 9 + lim (8 x + 72)

What is the tangent of (6x)/(x+8),\at x=1 ?
The tangent of (6x)/(x+8),\at x=1 is y= 16/27 x+2/27