What is the tangent of-2x^5+3x^2-7 ?

The tangent of-2x^5+3x^2-7 is y=(-10a_{0}^4+6a_{0})x+8a_{0}^5-3a_{0}^2-7

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tangent of-2x^5+3x^2-7

tangent of

- 2 x^{5} + 3 x^{2} - 7

y = (- 10 a_{0}^{4} + 6 a_{0}) x + 8 a_{0}^{5} - 3 a_{0}^{2} - 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 2 a_{0}^{5} + 3 a_{0}^{2} - 7)

Find the slope of

y = - 2 x^{5} + 3 x^{2} - 7 : \frac{d y}{d x} = - 10 x^{4} + 6 x

Compute the slope of

y = - 2 x^{5} + 3 x^{2} - 7

x = a_{0} : m = - 10 a_{0}^{4} + 6 a_{0}

Find the line with slope m=

- 10 a_{0}^{4} + 6 a_{0}

and passing through

(a_{0}, - 2 a_{0}^{5} + 3 a_{0}^{2} - 7) : y = (- 10 a_{0}^{4} + 6 a_{0}) x + 8 a_{0}^{5} - 3 a_{0}^{2} - 7

y = (- 10 a_{0}^{4} + 6 a_{0}) x + 8 a_{0}^{5} - 3 a_{0}^{2} - 7

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

\int (x + \frac{1}{x}) d x

\frac{d ^{2}}{d x ^{2}} (\frac{3 x - 2}{2 x + 5})

θ \to 0 lim (\frac{sin ( θ )}{θ})

\frac{d}{d x} (ln (x + 10))

What is the tangent of-2x^5+3x^2-7 ?
The tangent of-2x^5+3x^2-7 is y=(-10a_{0}^4+6a_{0})x+8a_{0}^5-3a_{0}^2-7