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

The tangent of y=x^4-4x^2+10,\at x=1 is y=-4x+11

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

tangent of

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

y = - 4 x + 11

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

y = x^{4} - 4 x^{2} + 10 : \frac{d y}{d x} = 4 x^{3} - 8 x

Compute the slope of

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

x = 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(1, 7) : y = - 4 x + 11

y = - 4 x + 11

\frac{d}{d x} (e^{- x} cos (2 x))

\int_{0}^{1} \frac{x}{1 - x ^{2}} d x

\frac{x d y}{d x} = 5 x e^{x} - y + 6 x^{2}

\int (x^{5} - 4 x)^{2} (5 x^{4} - 4) d x

x \to 0 + lim (\frac{1}{1 + \frac{1}{x}})

What is the tangent of y=x^4-4x^2+10,\at x=1 ?
The tangent of y=x^4-4x^2+10,\at x=1 is y=-4x+11