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

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

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

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

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 = - 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}, 7 + 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} + 7

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

\frac{d}{d x} ((e^{2 x}) d x)

\int (x^{3} e^{x^{4}}) d x

t^{2} \frac{d y}{d t} + y = 0

(tan^{c o s (x)} (x))^{^{'}}

y = (1 + 5 x)^{10}

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