What is the tangent of y=x^2,\at x= 1/3 ?

The tangent of y=x^2,\at x= 1/3 is y= 2/3 x-1/9

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tangent of y=x^2,\at x= 1/3

tangent of

y = x^{2}, at x = \frac{1}{3}

y = \frac{2}{3} x - \frac{1}{9}

Solution steps

Find the tangent point:

(\frac{1}{3}, \frac{1}{9})

Find the slope of

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

Find the line with slope m=

\frac{2}{3}

and passing through

(\frac{1}{3}, \frac{1}{9}) : y = \frac{2}{3} x - \frac{1}{9}

y = \frac{2}{3} x - \frac{1}{9}

\frac{\partial}{\partial y} (\frac{5 ( y + 2 )}{( x + 2 ) ^{2}})

\frac{\partial}{\partial x} (y^{x} - ln (x^{4} y + 6 x))

y^{^{''}} + y = - \frac{3}{4} sin (t) + \frac{1}{4} sin (3 t)

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

\frac{\partial}{\partial x} (sin (x) + cos (- 6 y))

What is the tangent of y=x^2,\at x= 1/3 ?
The tangent of y=x^2,\at x= 1/3 is y= 2/3 x-1/9