What is the tangent of x^4-13x^2+36 ?

The tangent of x^4-13x^2+36 is y=(4a_{0}^3-26a_{0})x-3a_{0}^4+13a_{0}^2+36

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tangent of x^4-13x^2+36

tangent of

x^{4} - 13 x^{2} + 36

y = (4 a_{0}^{3} - 26 a_{0}) x - 3 a_{0}^{4} + 13 a_{0}^{2} + 36

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 13 a_{0}^{2} + 36)

Find the slope of

y = x^{4} - 13 x^{2} + 36 : \frac{d y}{d x} = 4 x^{3} - 26 x

Compute the slope of

y = x^{4} - 13 x^{2} + 36

x = a_{0} : m = 4 a_{0}^{3} - 26 a_{0}

Find the line with slope m=

4 a_{0}^{3} - 26 a_{0}

and passing through

(a_{0}, a_{0}^{4} - 13 a_{0}^{2} + 36) : y = (4 a_{0}^{3} - 26 a_{0}) x - 3 a_{0}^{4} + 13 a_{0}^{2} + 36

y = (4 a_{0}^{3} - 26 a_{0}) x - 3 a_{0}^{4} + 13 a_{0}^{2} + 36

y = ln (8 x^{2} - 7 x + 9)

f (t) = e^{9 t s i n (2 t)}

y^{^{''}} + 49 y = sec^{2} (7 x)

\frac{x ^{2}}{x + 8}

\frac{\partial}{\partial x} (x^{3} y^{2} + 2 x^{2} y + x - 1)

What is the tangent of x^4-13x^2+36 ?
The tangent of x^4-13x^2+36 is y=(4a_{0}^3-26a_{0})x-3a_{0}^4+13a_{0}^2+36