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

The tangent of y=x^4+6x^2-x,(1,6) is y=15x-9

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tangent of y=x^4+6x^2-x,(1,6)

tangent of

y = x^{4} + 6 x^{2} - x, (1, 6)

y = 15 x - 9

Solution steps

Find the slope of

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

Compute the slope of

y = x^{4} + 6 x^{2} - x

(1, 6) : m = 15

Find the line with slope m=

15

and passing through

(1, 6) : y = 15 x - 9

y = 15 x - 9

\frac{\partial}{\partial y} (y^{x} - ln (x^{4} y + 7 x))

\frac{( 2 s + 3 )}{s ^{2} + 2 s + 26}

\int \frac{6 x}{1 - 7 x ^{2}} d x

2 x y^{^{'}} + y = 10 x

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

What is the tangent of y=x^4+6x^2-x,(1,6) ?
The tangent of y=x^4+6x^2-x,(1,6) is y=15x-9