What is the tangent of y=(x^{(3)}-4x)^{(12)}(2) ?

The tangent of y=(x^{(3)}-4x)^{(12)}(2) is

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tangent of y=(x^{(3)}-4x)^{(12)}(2)

tangent of

y = (x^{(3)} - 4 x)^{(12)} (2)

y = 24 (a_{0}^{3} - 4 a_{0})^{11} (3 a_{0}^{2} - 4) x - 70 a_{0}^{36} + 3168 a_{0}^{34} - 65472 a_{0}^{32} + 816640 a_{0}^{30} - 6842880 a_{0}^{28} + 40550400 a_{0}^{26} - 174096384 a_{0}^{24} + 544997376 a_{0}^{22} - 1232732160 a_{0}^{20} + 1960837120 a_{0}^{18} - 2076180480 a_{0}^{16} + 1308622848 a_{0}^{14} - 369098752 a_{0}^{12}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0}^{3} - 4 a_{0})^{12} \cdot 2)

Find the slope of

y = (x^{3} - 4 x)^{12} 2 : \frac{d y}{d x} = 24 (x^{3} - 4 x)^{11} (3 x^{2} - 4)

Compute the slope of

y = (x^{3} - 4 x)^{12} 2

x = a_{0} : m = 24 (a_{0}^{3} - 4 a_{0})^{11} (3 a_{0}^{2} - 4)

Find the line with slope m=

24 (a_{0}^{3} - 4 a_{0})^{11} (3 a_{0}^{2} - 4)

and passing through

(a_{0}, (a_{0}^{3} - 4 a_{0})^{12} 2) : y = 24 (a_{0}^{3} - 4 a_{0})^{11} (3 a_{0}^{2} - 4) x - 70 a_{0}^{36} + 3168 a_{0}^{34} - 65472 a_{0}^{32} + 816640 a_{0}^{30} - 6842880 a_{0}^{28} + 40550400 a_{0}^{26} - 174096384 a_{0}^{24} + 544997376 a_{0}^{22} - 1232732160 a_{0}^{20} + 1960837120 a_{0}^{18} - 2076180480 a_{0}^{16} + 1308622848 a_{0}^{14} - 369098752 a_{0}^{12}

y = 24 (a_{0}^{3} - 4 a_{0})^{11} (3 a_{0}^{2} - 4) x - 70 a_{0}^{36} + 3168 a_{0}^{34} - 65472 a_{0}^{32} + 816640 a_{0}^{30} - 6842880 a_{0}^{28} + 40550400 a_{0}^{26} - 174096384 a_{0}^{24} + 544997376 a_{0}^{22} - 1232732160 a_{0}^{20} + 1960837120 a_{0}^{18} - 2076180480 a_{0}^{16} + 1308622848 a_{0}^{14} - 369098752 a_{0}^{12}

\frac{\partial}{\partial y} (y + cos^{2} (x))

\int \frac{1}{x ^{2} - 6 x} d x

n = 1 \sum \infty tan (3 n)

g (x) = \frac{x ^{2} - 64}{x + 8}

y^{^{''}} + 6 y^{^{'}} + 5 y = - 3 t e^{(4 t)}

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