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

The tangent of y=(1+4x)^{3/2},\at x=2 is y=18x-9

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

tangent of

y = (1 + 4 x)^{\frac{3}{2}}, at x = 2

y = 18 x - 9

Solution steps

Find the tangent point:

(2, 27)

Find the slope of

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

Compute the slope of

y = (1 + 4 x)^{\frac{3}{2}}

x = 2 : m = 18

Find the line with slope m=

18

and passing through

(2, 27) : y = 18 x - 9

y = 18 x - 9

f (x) = 5 - 8 x^{2}, (- 2, - 27)

(x + 2 y) d x + (2 x - y) d y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0

x + y y^{^{'}} = 0

\frac{d y}{d x} = \frac{( e ^{x} )}{y ^{2}}

What is the tangent of y=(1+4x)^{3/2},\at x=2 ?
The tangent of y=(1+4x)^{3/2},\at x=2 is y=18x-9