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

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

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

tangent of

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

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

Solution steps

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

- \frac{4}{5}

and passing through

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

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

\frac{1}{x - 1}, 23

\int \frac{9}{5 - 6 x} d x

\frac{( 2 s + 1 )}{s ^{2} + 5 s + 6}

\frac{d y}{d t} = \frac{9}{40} - \frac{y}{200}

f (xg (x))

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