What is the tangent of x^3(1.1) ?

The tangent of x^3(1.1) is y=3.3a_{0}^2x-(11a_{0}^3)/5

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tangent of x^3(1.1)

tangent of

x^{3} (1.1)

y = 3.3 a_{0}^{2} x - \frac{11 a _{0}^{3}}{5}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} \cdot 1.1)

Find the slope of

y = x^{3} 1.1 : \frac{d y}{d x} = 3.3 x^{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 = 3.3 a_{0}^{2}

Find the line with slope m=

3.3 a_{0}^{2}

and passing through

(a_{0}, a_{0}^{3} 1.1) : y = 3.3 a_{0}^{2} x - \frac{11 a _{0}^{3}}{5}

y = 3.3 a_{0}^{2} x - \frac{11 a _{0}^{3}}{5}

\frac{\partial}{\partial x} (x z^{2} t)

\int \frac{x ^{5} - 6 x ^{3} - 6 x ^{2} - 8}{x ^{3} - 6 x} d x

\frac{y ^{^{'}}}{y ^{^{''}}} = \frac{1}{3}

\int (sec (x) + cot (x))^{2} d x

\frac{d}{d x} ((6 x^{2} + 5) (3 x + x))

What is the tangent of x^3(1.1) ?
The tangent of x^3(1.1) is y=3.3a_{0}^2x-(11a_{0}^3)/5