What is the tangent of y= x/(sqrt(81+x^2)),(0,0) ?

The tangent of y= x/(sqrt(81+x^2)),(0,0) is y= 1/9 x

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tangent of y= x/(sqrt(81+x^2)),(0,0)

tangent of

y = \frac{x}{81 + x ^{2}}, (0, 0)

y = \frac{1}{9} x

Solution steps

Find the slope of

y = \frac{x}{81 + x ^{2}} : \frac{d y}{d x} = \frac{81}{( 81 + x ^{2} ) 81 + x ^{2}}

Compute the slope of

y = \frac{x}{81 + x ^{2}}

(0, 0) : m = \frac{1}{9}

Find the line with slope m=

\frac{1}{9}

and passing through

(0, 0) : y = \frac{1}{9} x

y = \frac{1}{9} x

\int \frac{14}{x ( x ^{2} + 25 )} d x

y^{^{''}} - 4 y^{^{'}} - 5 y = 0

x + y^{2} = 42, x = y

f (x) = x - 2, at x = 6

\frac{d ^{2}}{d x ^{2}} (\frac{8}{9 t ^{5}})

What is the tangent of y= x/(sqrt(81+x^2)),(0,0) ?
The tangent of y= x/(sqrt(81+x^2)),(0,0) is y= 1/9 x