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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

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

\int \frac{1}{x ^{2} + 4 x + 5} d x

\int \frac{x ^{5} + x ^{2} + 2}{x ^{3} - x} d x

\int e^{- 3 x^{- 5}} x^{- 6} d x

\int sin^{3} (x) (cos^{2} (x)) d x

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