What is the tangent of 3x^4+x^3-21x^2 ?

The tangent of 3x^4+x^3-21x^2 is y=(12a_{0}^3+3a_{0}^2-42a_{0})x-9a_{0}^4-2a_{0}^3+21a_{0}^2

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tangent of 3x^4+x^3-21x^2

tangent of

3 x^{4} + x^{3} - 21 x^{2}

y = (12 a_{0}^{3} + 3 a_{0}^{2} - 42 a_{0}) x - 9 a_{0}^{4} - 2 a_{0}^{3} + 21 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{4} + a_{0}^{3} - 21 a_{0}^{2})

Find the slope of

y = 3 x^{4} + x^{3} - 21 x^{2} : \frac{d y}{d x} = 12 x^{3} + 3 x^{2} - 42 x

Compute the slope of

y = 3 x^{4} + x^{3} - 21 x^{2}

x = a_{0} : m = 12 a_{0}^{3} + 3 a_{0}^{2} - 42 a_{0}

Find the line with slope m=

12 a_{0}^{3} + 3 a_{0}^{2} - 42 a_{0}

and passing through

(a_{0}, 3 a_{0}^{4} + a_{0}^{3} - 21 a_{0}^{2}) : y = (12 a_{0}^{3} + 3 a_{0}^{2} - 42 a_{0}) x - 9 a_{0}^{4} - 2 a_{0}^{3} + 21 a_{0}^{2}

y = (12 a_{0}^{3} + 3 a_{0}^{2} - 42 a_{0}) x - 9 a_{0}^{4} - 2 a_{0}^{3} + 21 a_{0}^{2}

y = x (25.5)

7 x^{2}

y = (arcsin (sec (x)))^{t a n h (x)}

\int (a x + b) d x

What is the tangent of 3x^4+x^3-21x^2 ?
The tangent of 3x^4+x^3-21x^2 is y=(12a_{0}^3+3a_{0}^2-42a_{0})x-9a_{0}^4-2a_{0}^3+21a_{0}^2