What is the tangent of y=x^2+x ?

The tangent of y=x^2+x is y=(2a_{0}+1)x-a_{0}^2

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tangent of y=x^2+x

tangent of

y = x^{2} + x

y = (2 a_{0} + 1) x - a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = x^{2} + x : \frac{d y}{d x} = 2 x + 1

Compute the slope of

y = x^{2} + x

x = a_{0} : m = 2 a_{0} + 1

Find the line with slope m=

2 a_{0} + 1

and passing through

(a_{0}, a_{0}^{2} + a_{0}) : y = (2 a_{0} + 1) x - a_{0}^{2}

y = (2 a_{0} + 1) x - a_{0}^{2}

\int (x + y) d x

\frac{d y}{d x}, (3 x y + 7)^{2} = 6 y

\int sec^{2} (θ) tan^{2} (θ) d θ

y^{^{''}} + 62 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int_{- 1}^{1} 4 - x^{2} d x

What is the tangent of y=x^2+x ?
The tangent of y=x^2+x is y=(2a_{0}+1)x-a_{0}^2