What is the tangent of f(x)=x^4-20x^2+64 ?

The tangent of f(x)=x^4-20x^2+64 is y=(4a_{0}^3-40a_{0})x-3a_{0}^4+20a_{0}^2+64

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tangent of f(x)=x^4-20x^2+64

tangent of

f (x) = x^{4} - 20 x^{2} + 64

y = (4 a_{0}^{3} - 40 a_{0}) x - 3 a_{0}^{4} + 20 a_{0}^{2} + 64

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 20 a_{0}^{2} + 64)

Find the slope of

f (x) = x^{4} - 20 x^{2} + 64 : \frac{df ( x )}{d x} = 4 x^{3} - 40 x

Compute the slope of

f (x) = x^{4} - 20 x^{2} + 64

x = a_{0} : m = 4 a_{0}^{3} - 40 a_{0}

Find the line with slope m=

4 a_{0}^{3} - 40 a_{0}

and passing through

(a_{0}, a_{0}^{4} - 20 a_{0}^{2} + 64) : y = (4 a_{0}^{3} - 40 a_{0}) x - 3 a_{0}^{4} + 20 a_{0}^{2} + 64

y = (4 a_{0}^{3} - 40 a_{0}) x - 3 a_{0}^{4} + 20 a_{0}^{2} + 64

\frac{d}{d y} (sin (x y))

\int \frac{6 x + 5}{( 6 x ^{2} + 10 x ) ^{5}} d x

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

x \to 0 lim (- 3)

\frac{d}{d t} (e^{\frac{t}{7}} sin (\frac{t}{7}))

What is the tangent of f(x)=x^4-20x^2+64 ?
The tangent of f(x)=x^4-20x^2+64 is y=(4a_{0}^3-40a_{0})x-3a_{0}^4+20a_{0}^2+64