What is the tangent of y= x/(sqrt(36+x^2)),(0,0) ?

The tangent of y= x/(sqrt(36+x^2)),(0,0) is y= 1/6 x

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tangent of y= x/(sqrt(36+x^2)),(0,0)

tangent of

y = \frac{x}{36 + x ^{2}}, (0, 0)

y = \frac{1}{6} x

Solution steps

Find the slope of

y = \frac{x}{36 + x ^{2}} : \frac{d y}{d x} = \frac{36}{( 36 + x ^{2} ) 36 + x ^{2}}

Compute the slope of

y = \frac{x}{36 + x ^{2}}

(0, 0) : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(0, 0) : y = \frac{1}{6} x

y = \frac{1}{6} x

\int \frac{5}{x ( 7 + 5 x ) ^{6}} d x

f (t) = tan (sec (cos (t)))

x \to 1 + lim (x)

x \to 5 lim (76)

\frac{d y}{d x} = x + y, y (0) = 1

What is the tangent of y= x/(sqrt(36+x^2)),(0,0) ?
The tangent of y= x/(sqrt(36+x^2)),(0,0) is y= 1/6 x