What is the tangent of (x^2-9)^2 ?

The tangent of (x^2-9)^2 is y=4a_{0}(a_{0}^2-9)x-3a_{0}^4+18a_{0}^2+81

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tangent of (x^2-9)^2

tangent of

(x^{2} - 9)^{2}

y = 4 a_{0} (a_{0}^{2} - 9) x - 3 a_{0}^{4} + 18 a_{0}^{2} + 81

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0}^{2} - 9)^{2})

Find the slope of

y = (x^{2} - 9)^{2} : \frac{d y}{d x} = 4 x (x^{2} - 9)

Compute the slope of

y = (x^{2} - 9)^{2}

x = a_{0} : m = 4 a_{0} (a_{0}^{2} - 9)

Find the line with slope m=

4 a_{0} (a_{0}^{2} - 9)

and passing through

(a_{0}, (a_{0}^{2} - 9)^{2}) : y = 4 a_{0} (a_{0}^{2} - 9) x - 3 a_{0}^{4} + 18 a_{0}^{2} + 81

y = 4 a_{0} (a_{0}^{2} - 9) x - 3 a_{0}^{4} + 18 a_{0}^{2} + 81

y^{^{''}} - 3 y^{^{'}} - 10 y = 4, y (0) = 0, y^{^{'}} (0) = 0

\int_{1}^{e^{7}} \frac{ln ^{2} ( x ^{2} )}{x} d x

n = 1 \sum \infty (\frac{3}{n})^{4}

\int \frac{1}{sin ^{2} ( t )} d t

r = 1 0^{2 t}

What is the tangent of (x^2-9)^2 ?
The tangent of (x^2-9)^2 is y=4a_{0}(a_{0}^2-9)x-3a_{0}^4+18a_{0}^2+81