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}

\int \frac{3 x + 1}{4 - x ^{2}} d x

\int_{0}^{6} \frac{6 x}{1 + x ^{2}} d x

(x + 5) (7 x + 3)

y^{^{'}} - y = t

\frac{d}{d x} (x^{2} e^{x^{3}})