What is the tangent of f(x)=x^2 ?

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

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tangent of f(x)=x^2

tangent of

f (x) = x^{2}

y = 2 a_{0} 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})

Find the slope of

f (x) = x^{2} : \frac{df ( x )}{d x} = 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

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

y = 2 a_{0} x - a_{0}^{2}

2 x^{2} y \frac{d y}{d x} + 1 + 2 x y^{2} = 0

\frac{\partial}{\partial x} (x y + \frac{27}{x})

\frac{d}{d x} (e^{5 x + πi})

\int (6 + x^{5})^{2} 5 x^{4} d x

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