What is the tangent of y= x/(sqrt(25+x^2)),(0,0) ?

The tangent of y= x/(sqrt(25+x^2)),(0,0) is y= 1/5 x

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tangent of y= x/(sqrt(25+x^2)),(0,0)

tangent of

y = \frac{x}{25 + x ^{2}}, (0, 0)

y = \frac{1}{5} x

Solution steps

Find the slope of

y = \frac{x}{25 + x ^{2}} : \frac{d y}{d x} = \frac{25}{( 25 + x ^{2} ) 25 + x ^{2}}

Compute the slope of

y = \frac{x}{25 + x ^{2}}

(0, 0) : m = \frac{1}{5}

Find the line with slope m=

\frac{1}{5}

and passing through

(0, 0) : y = \frac{1}{5} x

y = \frac{1}{5} x

\int (4 e^{2 t} i + 3 e^{t} j) d t

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 2 y ^{2}} + 3 x)

x \to 1 lim (2 ∣ x - 1 ∣ - 1)

a y^{^{''}} + b y = 0

sin (\frac{π x}{2})

What is the tangent of y= x/(sqrt(25+x^2)),(0,0) ?
The tangent of y= x/(sqrt(25+x^2)),(0,0) is y= 1/5 x