What is the tangent of (6x)/(sqrt(4-x^2)),\at x=0 ?

The tangent of (6x)/(sqrt(4-x^2)),\at x=0 is y=3x

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tangent of (6x)/(sqrt(4-x^2)),\at x=0

tangent of

\frac{6 x}{4 - x ^{2}}, at x = 0

y = 3 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

y = \frac{6 x}{4 - x ^{2}} : \frac{d y}{d x} = \frac{24}{( 4 - x ^{2} ) 4 - x ^{2}}

Compute the slope of

y = \frac{6 x}{4 - x ^{2}}

x = 0 : m = 3

Find the line with slope m=

3

and passing through

(0, 0) : y = 3 x

y = 3 x

\frac{d}{d x} (4 7 x^{2} + 9)

\int \frac{x}{x ^{2} + 4} d x

5 x (x^{3} - 5 x + 5)

\frac{\partial}{\partial x} (4 ln (\frac{2 x y}{7 z}))

h (w) = \frac{4 w ^{5} - w}{w}

What is the tangent of (6x)/(sqrt(4-x^2)),\at x=0 ?
The tangent of (6x)/(sqrt(4-x^2)),\at x=0 is y=3x