What is the tangent of 1-x^2 ?

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

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tangent of 1-x^2

tangent of

1 - x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 2 a_{0}

Find the line with slope m=

- 2 a_{0}

and passing through

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

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

n = 1 \sum \infty n^{2} e^{- n}

x \to 0 lim (\frac{cos ( 5 x ) tan ( 5 x )}{3 x})

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

\frac{\partial}{\partial x} (sin^{2} (a x))

\int 5 e^{5 x} sin (e^{5 x}) d x