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

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

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tangent of x^2+3x

tangent of

x^{2} + 3 x

y = (2 a_{0} + 3) 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} + 3 a_{0})

Find the slope of

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

Compute the slope of

y = x^{2} + 3 x

x = a_{0} : m = 2 a_{0} + 3

Find the line with slope m=

2 a_{0} + 3

and passing through

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

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

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

\int_{l n (4)}^{l n (9)} e^{\frac{x}{2}} d x

f (x) = sin (\frac{1}{x ^{2}})

\int \frac{cos ^{6} ( x )}{sin ^{10} ( x )} d x

y^{^{''}} + \frac{a _{1^{2}} π ^{2}}{a _{2}^{2}} a_{3}^{2} y = 0