What is the tangent of y=8x^{1/2}+8x^{3/2}+6,\at x=4 ?

The tangent of y=8x^{1/2}+8x^{3/2}+6,\at x=4 is y=26x-18

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tangent of y=8x^{1/2}+8x^{3/2}+6,\at x=4

tangent of

y = 8 x^{\frac{1}{2}} + 8 x^{\frac{3}{2}} + 6, at x = 4

y = 26 x - 18

Solution steps

Find the tangent point:

(4, 86)

Find the slope of

y = 8 x^{\frac{1}{2}} + 8 x^{\frac{3}{2}} + 6 : \frac{d y}{d x} = \frac{4 + 12 x}{x ^{\frac{1}{2}}}

Compute the slope of

y = 8 x^{\frac{1}{2}} + 8 x^{\frac{3}{2}} + 6

x = 4 : m = 26

Find the line with slope m=

26

and passing through

(4, 86) : y = 26 x - 18

y = 26 x - 18

\int \frac{2 + x}{2 - x} d x

\frac{\partial}{\partial x} (\frac{1}{2 x ^{\frac{1}{2}}})

y = 20 x

y^{^{'''}} - 2 y^{^{''}} - y + 2 y = 0

x \to 0 lim (\frac{3 x ^{2} - 8 x}{x})

What is the tangent of y=8x^{1/2}+8x^{3/2}+6,\at x=4 ?
The tangent of y=8x^{1/2}+8x^{3/2}+6,\at x=4 is y=26x-18