What is the tangent of 4+5x^2-2x^3 ?

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

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tangent of 4+5x^2-2x^3

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 + 5 a_{0}^{2} - 2 a_{0}^{3})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = - 6 a_{0}^{2} + 10 a_{0}

Find the line with slope m=

- 6 a_{0}^{2} + 10 a_{0}

and passing through

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

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

\frac{d}{d x} (\frac{sin ( x ) + cos ( x )}{e ^{x}})

x \to 5 - lim (8 - x - x^{2})

y^{^{'}} = \frac{x - 2}{y - 2}

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} + 5 z^{2})

y = (\frac{( x + 2 )}{( x - 2 ) ^{2}}), at x = 3

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