What is the tangent of x^2+5x ?

The tangent of x^2+5x is y=(2a_{0}+5)x-a_{0}^2

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tangent of x^2+5x

tangent of

x^{2} + 5 x

y = (2 a_{0} + 5) 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} + 5 a_{0})

Find the slope of

y = x^{2} + 5 x : \frac{d y}{d x} = 2 x + 5

Compute the slope of

y = x^{2} + 5 x

x = a_{0} : m = 2 a_{0} + 5

Find the line with slope m=

2 a_{0} + 5

and passing through

(a_{0}, a_{0}^{2} + 5 a_{0}) : y = (2 a_{0} + 5) x - a_{0}^{2}

y = (2 a_{0} + 5) x - a_{0}^{2}

x \to 0 + lim (x^{\frac{1}{l n ( x )}})

y = x^{4} + 6 e^{x}, (0, 6)

5 \cdot t \cdot e^{- 5 \cdot t}

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

\frac{d}{d x} (x^{2} (x - 3))