What is the tangent of y=x^4-3x^2+10,\at x=1 ?

The tangent of y=x^4-3x^2+10,\at x=1 is y=-2x+10

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tangent of y=x^4-3x^2+10,\at x=1

tangent of

y = x^{4} - 3 x^{2} + 10, at x = 1

y = - 2 x + 10

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

y = x^{4} - 3 x^{2} + 10 : \frac{d y}{d x} = 4 x^{3} - 6 x

Compute the slope of

y = x^{4} - 3 x^{2} + 10

x = 1 : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 8) : y = - 2 x + 10

y = - 2 x + 10

x \cdot y^{^{'}} = y + 2

\int (y^{2} + 4 y)^{2} d y

5 x^{3} + 2

y^{^{'}} = y x e^{x + 2} - y

What is the tangent of y=x^4-3x^2+10,\at x=1 ?
The tangent of y=x^4-3x^2+10,\at x=1 is y=-2x+10