What is the tangent of-2x^2,\at 1 ?

The tangent of-2x^2,\at 1 is y=-4a_{0}x+2a_{0}^2

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tangent of-2x^2,\at 1

tangent of

- 2 x^{2}, at 1

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = - 2 x^{2} : \frac{d y}{d x} = - 4 x

Compute the slope of

y = - 2 x^{2}

x = a_{0} : m = - 4 a_{0}

Find the line with slope m=

- 4 a_{0}

and passing through

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

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

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

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

\int csc (x) cot^{2} (x) d x

\frac{d}{d x} (6^{2 x})

(e^{8 x})^{^{'}}

What is the tangent of-2x^2,\at 1 ?
The tangent of-2x^2,\at 1 is y=-4a_{0}x+2a_{0}^2