What is the tangent of (3x)/(4-x^2),\at x=1 ?

The tangent of (3x)/(4-x^2),\at x=1 is y= 5/3 x-2/3

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tangent of (3x)/(4-x^2),\at x=1

tangent of

\frac{3 x}{4 - x ^{2}}, at x = 1

y = \frac{5}{3} x - \frac{2}{3}

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

y = \frac{3 x}{4 - x ^{2}} : \frac{d y}{d x} = \frac{3 ( 4 + x ^{2} )}{( 4 - x ^{2} ) ^{2}}

Compute the slope of

y = \frac{3 x}{4 - x ^{2}}

x = 1 : m = \frac{5}{3}

Find the line with slope m=

\frac{5}{3}

and passing through

(1, 1) : y = \frac{5}{3} x - \frac{2}{3}

y = \frac{5}{3} x - \frac{2}{3}

\int_{b}^{1} 2 ln (x) d x

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

\frac{d}{d x} (\frac{x - x}{x ^{\frac{1}{5}}})

\int \frac{x ^{3}}{( 4 - x ^{2} ) ^{3}} d x

y^{^{'}} = (x + y - 1)^{2} - 1

What is the tangent of (3x)/(4-x^2),\at x=1 ?
The tangent of (3x)/(4-x^2),\at x=1 is y= 5/3 x-2/3