What is the tangent of 5/(4+x^2),\at x=2 ?

The tangent of 5/(4+x^2),\at x=2 is y=-5/16 x+5/4

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tangent of 5/(4+x^2),\at x=2

tangent of

\frac{5}{4 + x ^{2}}, at x = 2

y = - \frac{5}{16} x + \frac{5}{4}

Solution steps

Find the tangent point:

(2, \frac{5}{8})

Find the slope of

y = \frac{5}{4 + x ^{2}} : \frac{d y}{d x} = - \frac{10 x}{( 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}{16}

Find the line with slope m=

- \frac{5}{16}

and passing through

(2, \frac{5}{8}) : y = - \frac{5}{16} x + \frac{5}{4}

y = - \frac{5}{16} x + \frac{5}{4}

\int 3 csc^{2} (x) d x

y^{^{'}} + cot (x) y - x e^{x} = 0

\int (2 x + 1) e^{x (x + 1)} d x

d er i v a t i v e (10 x^{2} - 20 x + 20) e^{x}

d er i v a t i v e y = 4 e^{2 x}

What is the tangent of 5/(4+x^2),\at x=2 ?
The tangent of 5/(4+x^2),\at x=2 is y=-5/16 x+5/4