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

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

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tangent of x^3,\at x=1

tangent of

x^{3}, at x = 1

y = 3 x - 2

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

y = x^{3}

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 1) : y = 3 x - 2

y = 3 x - 2

y = (x^{6} - 10) x

\frac{d y}{d t} = 0.0572 (400 - y), y (0) = 32

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

\frac{\partial}{\partial y} (x y + yz + zy)

f (x) = - \frac{1}{9} (x^{- 9} - x^{18})