What is the tangent of x^4-8x^3+2 ?

The tangent of x^4-8x^3+2 is y=(4a_{0}^3-24a_{0}^2)x-3a_{0}^4+16a_{0}^3+2

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tangent of x^4-8x^3+2

tangent of

x^{4} - 8 x^{3} + 2

y = (4 a_{0}^{3} - 24 a_{0}^{2}) x - 3 a_{0}^{4} + 16 a_{0}^{3} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 8 a_{0}^{3} + 2)

Find the slope of

y = x^{4} - 8 x^{3} + 2 : \frac{d y}{d x} = 4 x^{3} - 24 x^{2}

Compute the slope of

y = x^{4} - 8 x^{3} + 2

x = a_{0} : m = 4 a_{0}^{3} - 24 a_{0}^{2}

Find the line with slope m=

4 a_{0}^{3} - 24 a_{0}^{2}

and passing through

(a_{0}, a_{0}^{4} - 8 a_{0}^{3} + 2) : y = (4 a_{0}^{3} - 24 a_{0}^{2}) x - 3 a_{0}^{4} + 16 a_{0}^{3} + 2

y = (4 a_{0}^{3} - 24 a_{0}^{2}) x - 3 a_{0}^{4} + 16 a_{0}^{3} + 2

\int sin (π x) d x

\int \frac{4 x ^{2} - 3 x - 4}{x} d x

- y d x + x d y = 0

\int \frac{1}{2 sin ( x ) - cos ( x )} d x

\frac{\partial}{\partial y} (cos (y))

What is the tangent of x^4-8x^3+2 ?
The tangent of x^4-8x^3+2 is y=(4a_{0}^3-24a_{0}^2)x-3a_{0}^4+16a_{0}^3+2