What is the tangent of f(x)= x/(x^2+16),\at x=0 ?

The tangent of f(x)= x/(x^2+16),\at x=0 is y= 1/16 x

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tangent of f(x)= x/(x^2+16),\at x=0

tangent of

f (x) = \frac{x}{x ^{2} + 16}, at x = 0

y = \frac{1}{16} x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = \frac{x}{x ^{2} + 16} : \frac{df ( x )}{d x} = \frac{- x ^{2} + 16}{( x ^{2} + 16 ) ^{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{1}{16}

Find the line with slope m=

\frac{1}{16}

and passing through

(0, 0) : y = \frac{1}{16} x

y = \frac{1}{16} x

\int \frac{x ^{2}}{( 15 + 4 x - 4 x ^{2} ) ^{\frac{3}{2}}} d x

\int (x^{2} e^{x^{3}} + x^{2} sin (y)) d x

\int (\frac{x - 5}{x ^{2} - x - 6}) d x

\int 7 sin (x) d x

\frac{d}{d x} (\frac{x + 3}{x ^{3} + x - 5})

What is the tangent of f(x)= x/(x^2+16),\at x=0 ?
The tangent of f(x)= x/(x^2+16),\at x=0 is y= 1/16 x