What is the tangent of y=8sqrt(x),(16,32) ?

The tangent of y=8sqrt(x),(16,32) is y=x+16

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tangent of y=8sqrt(x),(16,32)

tangent of

y = 8 x, (16, 32)

y = x + 16

Solution steps

Find the slope of

y = 8 x : \frac{d y}{d x} = \frac{4}{x}

Compute the slope of

y = 8 x

(16, 32) : m = 1

Find the line with slope m=

1

and passing through

(16, 32) : y = x + 16

y = x + 16

\frac{\partial}{\partial x} (x + \frac{1}{x})

\frac{\partial}{\partial x} ((x - 1)^{2} \cdot (y + 4)^{2})

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 t}

f (x) = x 3 - 2 x^{2}

y^{^{'}} = - y, y (0) = 1

What is the tangent of y=8sqrt(x),(16,32) ?
The tangent of y=8sqrt(x),(16,32) is y=x+16