What is the tangent of 9-x^2 ?

The tangent of 9-x^2 is y=-2a_{0}x+9+a_{0}^2

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tangent of 9-x^2

tangent of

9 - x^{2}

y = - 2 a_{0} x + 9 + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 - a_{0}^{2})

Find the slope of

y = 9 - x^{2} : \frac{d y}{d x} = - 2 x

Compute the slope of

y = 9 - x^{2}

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

Find the line with slope m=

- 2 a_{0}

and passing through

(a_{0}, 9 - a_{0}^{2}) : y = - 2 a_{0} x + 9 + a_{0}^{2}

y = - 2 a_{0} x + 9 + a_{0}^{2}

\frac{\partial}{\partial x} ((2 x + 6 y)^{11})

\frac{11 x}{( 2 + x ^{2} )}

x \to 2 lim (\frac{x + 1}{x - 1})

\frac{d}{d x} ((x + 5 x^{2})^{3})

y^{^{'}} = \frac{y ^{2}}{x ^{2}}