What is the tangent of 19-x^2 ?

The tangent of 19-x^2 is y=-2a_{0}x+19+a_{0}^2

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tangent of 19-x^2

tangent of

19 - x^{2}

y = - 2 a_{0} x + 19 + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 19 - a_{0}^{2})

Find the slope of

y = 19 - 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}, 19 - a_{0}^{2}) : y = - 2 a_{0} x + 19 + a_{0}^{2}

y = - 2 a_{0} x + 19 + a_{0}^{2}

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

y^{^{''}} = 4 y

\int tan (2 x) (sec (2 x))^{2} d x

\frac{s + 12}{( s ^{2} + 4 s )}

\frac{d}{d t} (\frac{5}{( 5 - t ) ^{2}})