What is the tangent of 4-x^2 ?

The tangent of 4-x^2 is y=-2a_{0}x+4+a_{0}^2

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tangent of 4-x^2

tangent of

4 - x^{2}

y = - 2 a_{0} x + 4 + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 - a_{0}^{2})

Find the slope of

y = 4 - 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}, 4 - a_{0}^{2}) : y = - 2 a_{0} x + 4 + a_{0}^{2}

y = - 2 a_{0} x + 4 + a_{0}^{2}

\frac{d y}{d x} = 2 x

\frac{d}{d x} (3 x + 3)

\int x^{\frac{1}{8}} + y^{\frac{1}{8}}

y^{^{''}} + k \cdot y = 0

\frac{\partial}{\partial x} (2 e^{4 x y})