What is the tangent of 33x^3 ?

The tangent of 33x^3 is y=99a_{0}^2x-66a_{0}^3

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tangent of 33x^3

tangent of

33 x^{3}

y = 99 a_{0}^{2} x - 66 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 33 a_{0}^{3})

Find the slope of

y = 33 x^{3} : \frac{d y}{d x} = 99 x^{2}

Compute the slope of

y = 33 x^{3}

x = a_{0} : m = 99 a_{0}^{2}

Find the line with slope m=

99 a_{0}^{2}

and passing through

(a_{0}, 33 a_{0}^{3}) : y = 99 a_{0}^{2} x - 66 a_{0}^{3}

y = 99 a_{0}^{2} x - 66 a_{0}^{3}

\int \frac{a + b x ^{7}}{8 a x + b x ^{8}} d x

f (x) = 4 e^{x} + x, at x = 0

\frac{d}{d x} (cos (\frac{1 - e ^{7 x}}{1 + e ^{7 x}}))

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

f (x) = x^{2} \cdot e^{x}