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

The tangent of x^5-4x^3+2x-3 is y=(5a_{0}^4-12a_{0}^2+2)x-4a_{0}^5+8a_{0}^3-3

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

tangent of

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

y = (5 a_{0}^{4} - 12 a_{0}^{2} + 2) x - 4 a_{0}^{5} + 8 a_{0}^{3} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = x^{5} - 4 x^{3} + 2 x - 3 : \frac{d y}{d x} = 5 x^{4} - 12 x^{2} + 2

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 5 a_{0}^{4} - 12 a_{0}^{2} + 2

Find the line with slope m=

5 a_{0}^{4} - 12 a_{0}^{2} + 2

and passing through

(a_{0}, a_{0}^{5} - 4 a_{0}^{3} + 2 a_{0} - 3) : y = (5 a_{0}^{4} - 12 a_{0}^{2} + 2) x - 4 a_{0}^{5} + 8 a_{0}^{3} - 3

y = (5 a_{0}^{4} - 12 a_{0}^{2} + 2) x - 4 a_{0}^{5} + 8 a_{0}^{3} - 3

\frac{d}{d y} (y^{2} e^{x y^{2}} + 4 x^{3})

\frac{7 x ^{2} + 5 x}{4 x}

\int \frac{lo g _{9} ( x )}{x} d x

f (x) = \frac{3 x + 2}{x - 1}

\frac{\partial}{\partial x} (81 x^{8} y^{8} - 64 x^{7} y^{7})

What is the tangent of x^5-4x^3+2x-3 ?
The tangent of x^5-4x^3+2x-3 is y=(5a_{0}^4-12a_{0}^2+2)x-4a_{0}^5+8a_{0}^3-3