What is the tangent of 3/(x+3),\at x=4 ?

The tangent of 3/(x+3),\at x=4 is y=-3/49 x+33/49

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tangent of 3/(x+3),\at x=4

tangent of

\frac{3}{x + 3}, at x = 4

y = - \frac{3}{49} x + \frac{33}{49}

Solution steps

Find the tangent point:

(4, \frac{3}{7})

Find the slope of

y = \frac{3}{x + 3} : \frac{d y}{d x} = - \frac{3}{( x + 3 ) ^{2}}

Compute the slope of

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

x = 4 : m = - \frac{3}{49}

Find the line with slope m=

- \frac{3}{49}

and passing through

(4, \frac{3}{7}) : y = - \frac{3}{49} x + \frac{33}{49}

y = - \frac{3}{49} x + \frac{33}{49}

y^{^{'}} = \frac{( x ^{3} + y ^{3} )}{3 x y ^{2}}

\frac{y}{x ^{2}} d x + (y - \frac{1}{x}) d y = 0

\frac{\partial}{\partial x} (e^{5 x yz - 9})

\frac{\partial}{\partial y} (\frac{1}{y + 2 z})

f (t) = t^{2} - t + 1

What is the tangent of 3/(x+3),\at x=4 ?
The tangent of 3/(x+3),\at x=4 is y=-3/49 x+33/49