What is the tangent of y= 3/(sqrt(16-x)),\at x=0 ?

The tangent of y= 3/(sqrt(16-x)),\at x=0 is y= 3/128 x+3/4

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tangent of y= 3/(sqrt(16-x)),\at x=0

tangent of

y = \frac{3}{16 - x}, at x = 0

y = \frac{3}{128} x + \frac{3}{4}

Solution steps

Find the tangent point:

(0, \frac{3}{4})

Find the slope of

y = \frac{3}{16 - x} : \frac{d y}{d x} = \frac{3}{2 ( 16 - x ) ^{\frac{3}{2}}}

Compute the slope of

y = \frac{3}{16 - x}

x = 0 : m = \frac{3}{128}

Find the line with slope m=

\frac{3}{128}

and passing through

(0, \frac{3}{4}) : y = \frac{3}{128} x + \frac{3}{4}

y = \frac{3}{128} x + \frac{3}{4}

\int x e^{- x (y + 1)} d x

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

\int_{0}^{\frac{1}{2}} 1 d x

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

\frac{d y}{d x} + y cos (x) = 2 cos (x), y (0) = 4

What is the tangent of y= 3/(sqrt(16-x)),\at x=0 ?
The tangent of y= 3/(sqrt(16-x)),\at x=0 is y= 3/128 x+3/4