What is the tangent of f(x)= 1/(2+3x),\at x=1 ?

The tangent of f(x)= 1/(2+3x),\at x=1 is y=-3/25 x+8/25

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tangent of f(x)= 1/(2+3x),\at x=1

tangent of

f (x) = \frac{1}{2 + 3 x}, at x = 1

y = - \frac{3}{25} x + \frac{8}{25}

Solution steps

Find the tangent point:

(1, \frac{1}{5})

Find the slope of

f (x) = \frac{1}{2 + 3 x} : \frac{df ( x )}{d x} = - \frac{3}{( 2 + 3 x ) ^{2}}

Compute the slope of

f (x) = \frac{1}{2 + 3 x}

x = 1 : m = - \frac{3}{25}

Find the line with slope m=

- \frac{3}{25}

and passing through

(1, \frac{1}{5}) : y = - \frac{3}{25} x + \frac{8}{25}

y = - \frac{3}{25} x + \frac{8}{25}

y = - 3 x - 5

\frac{\partial}{\partial x} (- 2 (y + 1) x^{- 3})

sin^{2} (x) - cos^{2} (x)

2 sin (x) y^{^{'}} + y cos (x) = y^{2} cos (x)

\frac{\partial}{\partial s} (r sin (t) sin (s))

What is the tangent of f(x)= 1/(2+3x),\at x=1 ?
The tangent of f(x)= 1/(2+3x),\at x=1 is y=-3/25 x+8/25