What is the tangent of-x^2+1 ?

The tangent of-x^2+1 is y=-2a_{0}x+a_{0}^2+1

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tangent of-x^2+1

tangent of

- x^{2} + 1

y = - 2 a_{0} x + a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - a_{0}^{2} + 1)

Find the slope of

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

Compute the slope of

y = - x^{2} + 1

x = a_{0} : m = - 2 a_{0}

Find the line with slope m=

- 2 a_{0}

and passing through

(a_{0}, - a_{0}^{2} + 1) : y = - 2 a_{0} x + a_{0}^{2} + 1

y = - 2 a_{0} x + a_{0}^{2} + 1

(x^{3} - 4 x) (5 x^{2} + 2 x)

\frac{d}{d x} (e^{x^{4} + 7 x})

\frac{\partial}{\partial x} (7 + x ln (x y - 9))

\frac{d}{d x} ((x^{3} + 5 x^{2} - x^{- 2} + 8)^{- \frac{7}{6}})

q \to 0 + lim (4 q)