What is the tangent of f(x)=10-x^2 ?

The tangent of f(x)=10-x^2 is y=-2a_{0}x+10+a_{0}^2

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tangent of f(x)=10-x^2

tangent of

f (x) = 10 - x^{2}

y = - 2 a_{0} x + 10 + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 10 - a_{0}^{2})

Find the slope of

f (x) = 10 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 10 - x^{2}

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

Find the line with slope m=

- 2 a_{0}

and passing through

(a_{0}, 10 - a_{0}^{2}) : y = - 2 a_{0} x + 10 + a_{0}^{2}

y = - 2 a_{0} x + 10 + a_{0}^{2}

7 x^{2} \frac{d y}{d x} = \frac{d y}{d x} + 7 x e^{- y}

\int sin (\frac{x}{2}) \cdot cos (x) d x

\frac{d}{d x} (x^{2} - \frac{1}{x ^{2}})

\frac{1}{x ^{2}}, 3

t^{2} y^{^{''}} + 4 t y^{^{'}} - 18 y = 0

What is the tangent of f(x)=10-x^2 ?
The tangent of f(x)=10-x^2 is y=-2a_{0}x+10+a_{0}^2