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

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

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

tangent of

f (x) = (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

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

Compute the slope of

f (x) = (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

y = x^{2} - 5 x - 3

\frac{d}{d x} (\frac{tan ^{2} ( x )}{2})

\frac{d}{d x} (e^{3 x + 3})

\int \frac{1}{2 x ^{2} - 5 x + 7} d x

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

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