What is the tangent of x^2,\at x=2 ?

The tangent of x^2,\at x=2 is y=4x-4

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tangent of x^2,\at x=2

tangent of

x^{2}, at x = 2

y = 4 x - 4

Solution steps

Find the tangent point:

(2, 4)

Find the slope of

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

Compute the slope of

y = x^{2}

x = 2 : m = 4

Find the line with slope m=

4

and passing through

(2, 4) : y = 4 x - 4

y = 4 x - 4

\frac{\partial}{\partial x} (\frac{4 - x ^{2} y}{2 x})

\frac{\partial}{\partial x} (3 x y^{2} + 20 x^{2} y^{3})

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{4}) = 2, y^{^{'}} (\frac{π}{4}) = - 2

\int \frac{3 x ^{2} + 17 x + 32}{x ^{3} + 8 x ^{2} + 16 x} d x

y^{^{''}} = 1 + (y^{^{'}})^{2}