What is the tangent of f(x)=(x^2-4)^2 ?

The tangent of f(x)=(x^2-4)^2 is y=4a_{0}(a_{0}^2-4)x-3a_{0}^4+8a_{0}^2+16

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tangent of f(x)=(x^2-4)^2

tangent of

f (x) = (x^{2} - 4)^{2}

y = 4 a_{0} (a_{0}^{2} - 4) x - 3 a_{0}^{4} + 8 a_{0}^{2} + 16

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = (x^{2} - 4)^{2} : \frac{df ( x )}{d x} = 4 x (x^{2} - 4)

Compute the slope of

f (x) = (x^{2} - 4)^{2}

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

Find the line with slope m=

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

and passing through

(a_{0}, (a_{0}^{2} - 4)^{2}) : y = 4 a_{0} (a_{0}^{2} - 4) x - 3 a_{0}^{4} + 8 a_{0}^{2} + 16

y = 4 a_{0} (a_{0}^{2} - 4) x - 3 a_{0}^{4} + 8 a_{0}^{2} + 16

\frac{d}{d t} (a cos (t))

\frac{3}{( s + 1 ) ^{3}}

f (x) = \frac{x}{1 + x}

f (x) = 4 x^{3}

\frac{d x}{d t} = x + e^{t}

What is the tangent of f(x)=(x^2-4)^2 ?
The tangent of f(x)=(x^2-4)^2 is y=4a_{0}(a_{0}^2-4)x-3a_{0}^4+8a_{0}^2+16