What is the tangent of f(x)=x^4-10x^2+9 ?

The tangent of f(x)=x^4-10x^2+9 is y=(4a_{0}^3-20a_{0})x-3a_{0}^4+10a_{0}^2+9

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tangent of f(x)=x^4-10x^2+9

tangent of

f (x) = x^{4} - 10 x^{2} + 9

y = (4 a_{0}^{3} - 20 a_{0}) x - 3 a_{0}^{4} + 10 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 10 a_{0}^{2} + 9)

Find the slope of

f (x) = x^{4} - 10 x^{2} + 9 : \frac{df ( x )}{d x} = 4 x^{3} - 20 x

Compute the slope of

f (x) = x^{4} - 10 x^{2} + 9

x = a_{0} : m = 4 a_{0}^{3} - 20 a_{0}

Find the line with slope m=

4 a_{0}^{3} - 20 a_{0}

and passing through

(a_{0}, a_{0}^{4} - 10 a_{0}^{2} + 9) : y = (4 a_{0}^{3} - 20 a_{0}) x - 3 a_{0}^{4} + 10 a_{0}^{2} + 9

y = (4 a_{0}^{3} - 20 a_{0}) x - 3 a_{0}^{4} + 10 a_{0}^{2} + 9

\int - \frac{14}{x ^{15}} d x

\int x^{n} d x

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

x \to 0 - lim (sin (\frac{1}{x}))

\frac{d}{d x} (e^{x} (3 x^{2} + 5 x - 1))

What is the tangent of f(x)=x^4-10x^2+9 ?
The tangent of f(x)=x^4-10x^2+9 is y=(4a_{0}^3-20a_{0})x-3a_{0}^4+10a_{0}^2+9