What is the tangent of-10x^4+x^3 ?

The tangent of-10x^4+x^3 is y=(-40a_{0}^3+3a_{0}^2)x+30a_{0}^4-2a_{0}^3

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tangent of-10x^4+x^3

tangent of

- 10 x^{4} + x^{3}

y = (- 40 a_{0}^{3} + 3 a_{0}^{2}) x + 30 a_{0}^{4} - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 10 a_{0}^{4} + a_{0}^{3})

Find the slope of

y = - 10 x^{4} + x^{3} : \frac{d y}{d x} = - 40 x^{3} + 3 x^{2}

Compute the slope of

y = - 10 x^{4} + x^{3}

x = a_{0} : m = - 40 a_{0}^{3} + 3 a_{0}^{2}

Find the line with slope m=

- 40 a_{0}^{3} + 3 a_{0}^{2}

and passing through

(a_{0}, - 10 a_{0}^{4} + a_{0}^{3}) : y = (- 40 a_{0}^{3} + 3 a_{0}^{2}) x + 30 a_{0}^{4} - 2 a_{0}^{3}

y = (- 40 a_{0}^{3} + 3 a_{0}^{2}) x + 30 a_{0}^{4} - 2 a_{0}^{3}

y = x^{\frac{13}{14}}

n = 1 \sum \infty ln ((n + 1)^{2})

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 4

n = 0 \sum \infty \frac{1}{3 ^{n + 1}}

\int tan^{3} (2 t) sec^{3} (2 t) d t

What is the tangent of-10x^4+x^3 ?
The tangent of-10x^4+x^3 is y=(-40a_{0}^3+3a_{0}^2)x+30a_{0}^4-2a_{0}^3