What is the tangent of y=x^3-2x+1,(4,57) ?

The tangent of y=x^3-2x+1,(4,57) is y=46x-127

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tangent of y=x^3-2x+1,(4,57)

tangent of

y = x^{3} - 2 x + 1, (4, 57)

y = 46 x - 127

Solution steps

Find the slope of

y = x^{3} - 2 x + 1 : \frac{d y}{d x} = 3 x^{2} - 2

Compute the slope of

y = x^{3} - 2 x + 1

(4, 57) : m = 46

Find the line with slope m=

46

and passing through

(4, 57) : y = 46 x - 127

y = 46 x - 127

f (x) = \frac{sin ( x )}{1 + cos ( x )}

\frac{d}{d x} (\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} - 1)

y^{^{'}} = \frac{x}{6 y}

\frac{d y}{d t} = 0.12 - \frac{6 y}{50 - 2 t}

\int x sin (x + 1) d x

What is the tangent of y=x^3-2x+1,(4,57) ?
The tangent of y=x^3-2x+1,(4,57) is y=46x-127