What is the tangent of 2y^2-sqrt(x)=14,(16,3) ?

The tangent of 2y^2-sqrt(x)=14,(16,3) is y= 1/96 x+17/6

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tangent of 2y^2-sqrt(x)=14,(16,3)

tangent of

2 y^{2} - x = 14, (16, 3)

y = \frac{1}{96} x + \frac{17}{6}

Solution steps

Find the slope of

2 y^{2} - x = 14 : \frac{d y}{d x} = \frac{1}{8 y x}

Compute the slope of

2 y^{2} - x = 14

(16, 3) : m = \frac{1}{96}

Find the line with slope m=

\frac{1}{96}

and passing through

(16, 3) : y = \frac{1}{96} x + \frac{17}{6}

y = \frac{1}{96} x + \frac{17}{6}

y = ln (\frac{1}{x x + 4})

f (x) = x^{3}

(- 43, - 4)

x^{2} e^{x}

e^{- 2 x}

What is the tangent of 2y^2-sqrt(x)=14,(16,3) ?
The tangent of 2y^2-sqrt(x)=14,(16,3) is y= 1/96 x+17/6