What is the tangent of y=x^4+3x^2-x,(1,3) ?

The tangent of y=x^4+3x^2-x,(1,3) is y=9x-6

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tangent of y=x^4+3x^2-x,(1,3)

tangent of

y = x^{4} + 3 x^{2} - x, (1, 3)

y = 9 x - 6

Solution steps

Find the slope of

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

Compute the slope of

y = x^{4} + 3 x^{2} - x

(1, 3) : m = 9

Find the line with slope m=

9

and passing through

(1, 3) : y = 9 x - 6

y = 9 x - 6

\frac{\partial}{\partial x} (2 y - 2 y x)

\frac{d}{d x} (3 ln (sin (x)))

\frac{d ^{2}}{d t ^{2}} (- 9.8 t^{2} + 2 t + 1)

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

(x + 8)^{2}

What is the tangent of y=x^4+3x^2-x,(1,3) ?
The tangent of y=x^4+3x^2-x,(1,3) is y=9x-6