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

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

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

tangent of

\frac{4}{5 x + 1}, at x = - 1

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

Solution steps

Find the tangent point:

(- 1, - 1)

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}

x = - 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}

x \to 3 lim (3 - x)

x \to \infty lim (8 x)

x y^{^{'}} + 2 y = \frac{2}{x ^{2}} + 1

\int x^{3} e x^{2} d x

\int (n x) \frac{1 - n}{n} d x

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