What is the tangent of y=x^4-5x^3+2,\at x=2 ?

The tangent of y=x^4-5x^3+2,\at x=2 is y=-28x+34

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tangent of y=x^4-5x^3+2,\at x=2

tangent of

y = x^{4} - 5 x^{3} + 2, at x = 2

y = - 28 x + 34

Solution steps

Find the tangent point:

(2, - 22)

Find the slope of

y = x^{4} - 5 x^{3} + 2 : \frac{d y}{d x} = 4 x^{3} - 15 x^{2}

Compute the slope of

y = x^{4} - 5 x^{3} + 2

x = 2 : m = - 28

Find the line with slope m=

- 28

and passing through

(2, - 22) : y = - 28 x + 34

y = - 28 x + 34

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

\frac{\partial}{\partial x} (ln (x^{2} y))

x \to 0 lim (x sin (\frac{3 π}{x}))

y = - \frac{14}{x ^{5}}

\frac{\partial}{\partial x} (6 x^{3} cos (6 x^{2} y))

What is the tangent of y=x^4-5x^3+2,\at x=2 ?
The tangent of y=x^4-5x^3+2,\at x=2 is y=-28x+34