What is the tangent of y=(6x^2-6x+2)(1+2x),\at x=1 ?

The tangent of y=(6x^2-6x+2)(1+2x),\at x=1 is y=22x-16

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tangent of y=(6x^2-6x+2)(1+2x),\at x=1

tangent of

y = (6 x^{2} - 6 x + 2) (1 + 2 x), at x = 1

y = 22 x - 16

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

y = (6 x^{2} - 6 x + 2) (1 + 2 x) : \frac{d y}{d x} = 36 x^{2} - 12 x - 2

Compute the slope of

y = (6 x^{2} - 6 x + 2) (1 + 2 x)

x = 1 : m = 22

Find the line with slope m=

22

and passing through

(1, 6) : y = 22 x - 16

y = 22 x - 16

\frac{\partial}{\partial y} (cos (x^{2} + y^{3}))

sin^{2} (7 t)

\frac{d}{d x} (- \frac{1}{3})

\int cos^{16} (x) sin (x) d x

\frac{d}{d x} ((x^{2} - 1)^{4} (2 - 3 x))

What is the tangent of y=(6x^2-6x+2)(1+2x),\at x=1 ?
The tangent of y=(6x^2-6x+2)(1+2x),\at x=1 is y=22x-16