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

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

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

tangent of

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

y^{^{'}} + x y = 0

x \to 0 + lim (cot^{2} (x))

\frac{d}{d x} (x^{4} - 9 x + 6)

\frac{d}{d x} (a tan (x))

\int \frac{t ^{2}}{a ^{4} + t ^{2}} d t

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