What is the tangent of y= 4/(5x+1),(-1,-1) ?

The tangent of y= 4/(5x+1),(-1,-1) is y=-5/4 x-9/4

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tangent of y= 4/(5x+1),(-1,-1)

tangent of

y = \frac{4}{5 x + 1}, (- 1, - 1)

y = - \frac{5}{4} x - \frac{9}{4}

Solution steps

Find the slope of

y = \frac{4}{5 x + 1} : \frac{d y}{d x} = - \frac{20}{( 5 x + 1 ) ^{2}}

Compute the slope of

y = \frac{4}{5 x + 1}

(- 1, - 1) : m = - \frac{5}{4}

Find the line with slope m=

- \frac{5}{4}

and passing through

(- 1, - 1) : y = - \frac{5}{4} x - \frac{9}{4}

y = - \frac{5}{4} x - \frac{9}{4}

\frac{9}{x ^{2}} - \frac{3}{x}

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

\int_{- \infty}^{6} \frac{1}{x ^{2} + 1} d x

\frac{d}{d x} (2 e^{- x^{3}})

\int - \frac{1}{v ^{2}} d v

What is the tangent of y= 4/(5x+1),(-1,-1) ?
The tangent of y= 4/(5x+1),(-1,-1) is y=-5/4 x-9/4