What is the tangent of (2x^2+1)3(x^2-4)2 ?

The tangent of (2x^2+1)3(x^2-4)2 is y=6(8a_{0}^3-14a_{0})x-36a_{0}^4+42a_{0}^2-24

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tangent of (2x^2+1)3(x^2-4)2

tangent of

(2 x^{2} + 1) 3 (x^{2} - 4) 2

y = 6 (8 a_{0}^{3} - 14 a_{0}) x - 36 a_{0}^{4} + 42 a_{0}^{2} - 24

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 (2 a_{0}^{2} + 1) (a_{0}^{2} - 4))

Find the slope of

y = (2 x^{2} + 1) 3 (x^{2} - 4) 2 : \frac{d y}{d x} = 6 (8 x^{3} - 14 x)

Compute the slope of

y = (2 x^{2} + 1) 3 (x^{2} - 4) 2

x = a_{0} : m = 6 (8 a_{0}^{3} - 14 a_{0})

Find the line with slope m=

6 (8 a_{0}^{3} - 14 a_{0})

and passing through

(a_{0}, 6 (2 a_{0}^{2} + 1) (a_{0}^{2} - 4)) : y = 6 (8 a_{0}^{3} - 14 a_{0}) x - 36 a_{0}^{4} + 42 a_{0}^{2} - 24

y = 6 (8 a_{0}^{3} - 14 a_{0}) x - 36 a_{0}^{4} + 42 a_{0}^{2} - 24

\int \frac{cos ^{2} ( x ) sin ( x )}{2} d x

x \to π lim (ln (cos (x) - 4))

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (2 t)

\int \frac{2 x + 3}{( x ^{2} + 3 x + 1 ) ^{3}} d x

\int \frac{ln ( x ^{18} )}{x} d x

What is the tangent of (2x^2+1)3(x^2-4)2 ?
The tangent of (2x^2+1)3(x^2-4)2 is y=6(8a_{0}^3-14a_{0})x-36a_{0}^4+42a_{0}^2-24