What is the tangent of y=x^3-2x+2,(3,23) ?

The tangent of y=x^3-2x+2,(3,23) is y=25x-52

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tangent of y=x^3-2x+2,(3,23)

tangent of

y = x^{3} - 2 x + 2, (3, 23)

y = 25 x - 52

Solution steps

Find the slope of

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

Compute the slope of

y = x^{3} - 2 x + 2

(3, 23) : m = 25

Find the line with slope m=

25

and passing through

(3, 23) : y = 25 x - 52

y = 25 x - 52

\frac{d y}{d x} = \frac{( 3 y ^{2} )}{x}

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

6

\frac{\partial}{\partial x} (4 x^{4} y^{8} + 9 x^{6} y^{7})

\frac{d}{d x} (2^{x} (x^{2} + 1))

What is the tangent of y=x^3-2x+2,(3,23) ?
The tangent of y=x^3-2x+2,(3,23) is y=25x-52