What is the tangent of x^2+x-3 ?

The tangent of x^2+x-3 is y=(2a_{0}+1)x-a_{0}^2-3

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tangent of x^2+x-3

tangent of

x^{2} + x - 3

y = (2 a_{0} + 1) x - a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + a_{0} - 3)

Find the slope of

y = x^{2} + x - 3 : \frac{d y}{d x} = 2 x + 1

Compute the slope of

y = x^{2} + x - 3

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} - 3) : y = (2 a_{0} + 1) x - a_{0}^{2} - 3

y = (2 a_{0} + 1) x - a_{0}^{2} - 3

\frac{\partial}{\partial y} (3 x^{2} y^{4} - 3)

f (x) = 4 x^{3}

\frac{d}{d x} (ln (1 + x y))

\int 30 (x + 7) (x + 5)^{4} d x

\frac{d}{d x} (ln (\frac{x}{4}))

What is the tangent of x^2+x-3 ?
The tangent of x^2+x-3 is y=(2a_{0}+1)x-a_{0}^2-3