What is the tangent of y= x/(sqrt(9+x^2)),(0,0) ?

The tangent of y= x/(sqrt(9+x^2)),(0,0) is y= 1/3 x

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tangent of y= x/(sqrt(9+x^2)),(0,0)

tangent of

y = \frac{x}{9 + x ^{2}}, (0, 0)

y = \frac{1}{3} x

Solution steps

Find the slope of

y = \frac{x}{9 + x ^{2}} : \frac{d y}{d x} = \frac{9}{( 9 + x ^{2} ) 9 + x ^{2}}

Compute the slope of

y = \frac{x}{9 + x ^{2}}

(0, 0) : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(0, 0) : y = \frac{1}{3} x

y = \frac{1}{3} x

\int 40 sin (\frac{π x}{3}) + 50 d x

\int 3 + x \cdot (x + 1)^{2} d x

\frac{\partial}{\partial x} (x^{\frac{1}{2}})

\frac{d}{d x} (\frac{- 8 x}{x ^{2} + 1})

y = 3 x^{2} ln (x), y = 12 ln (x)

What is the tangent of y= x/(sqrt(9+x^2)),(0,0) ?
The tangent of y= x/(sqrt(9+x^2)),(0,0) is y= 1/3 x