What is the tangent of x^2-sqrt(x) ?

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

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tangent of x^2-sqrt(x)

tangent of

x^{2} - x

y = (2 a_{0} - \frac{1}{2 a _{0}}) x - a_{0}^{2} - a_{0} + \frac{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 - \frac{1}{2 x}

Compute the slope of

y = x^{2} - x

x = a_{0} : m = 2 a_{0} - \frac{1}{2 a _{0}}

Find the line with slope m=

2 a_{0} - \frac{1}{2 a _{0}}

and passing through

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

y = (2 a_{0} - \frac{1}{2 a _{0}}) x - a_{0}^{2} - a_{0} + \frac{a _{0}}{2}

x \to 0 lim (\frac{∣ x ∣}{2 x})

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

\int \frac{5}{x ^{2} - 1} d x

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

x \to 2 lim (\frac{1}{x ^{2}})

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