What is the tangent of (6x)/(x+7),\at x=1 ?

The tangent of (6x)/(x+7),\at x=1 is y= 21/32 x+3/32

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tangent of (6x)/(x+7),\at x=1

tangent of

\frac{6 x}{x + 7}, at x = 1

y = \frac{21}{32} x + \frac{3}{32}

Solution steps

Find the tangent point:

(1, \frac{3}{4})

Find the slope of

y = \frac{6 x}{x + 7} : \frac{d y}{d x} = \frac{42}{( x + 7 ) ^{2}}

Compute the slope of

y = \frac{6 x}{x + 7}

x = 1 : m = \frac{21}{32}

Find the line with slope m=

\frac{21}{32}

and passing through

(1, \frac{3}{4}) : y = \frac{21}{32} x + \frac{3}{32}

y = \frac{21}{32} x + \frac{3}{32}

\frac{\partial}{\partial y} (x e^{\frac{y}{x}})

x^{2} - 2

\int \frac{y ^{2} + 6 y + 9}{( y ^{2} + 9 ) ^{2}} d y

f (x) = 3 x - 1, at x = 5

n = 1 \sum \infty \frac{n !}{( 2 n ) !}

What is the tangent of (6x)/(x+7),\at x=1 ?
The tangent of (6x)/(x+7),\at x=1 is y= 21/32 x+3/32