What is the tangent of y=ax^3 ?

The tangent of y=ax^3 is y=3aa_{1}^2x-2aa_{1}^3

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tangent of y=ax^3

tangent of

y = a x^{3}

y = 3 a a_{1}^{2} x - 2 a a_{1}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{1}

Find the tangent point:

(a_{1}, a a_{1}^{3})

Find the slope of

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

Compute the slope of

y = a x^{3}

x = a_{1} : m = 3 a a_{1}^{2}

Find the line with slope m=

3 a a_{1}^{2}

and passing through

(a_{1}, a a_{1}^{3}) : y = 3 a a_{1}^{2} x - 2 a a_{1}^{3}

y = 3 a a_{1}^{2} x - 2 a a_{1}^{3}

\int_{0}^{1} \frac{ln ( t )}{t} d t

\int + x^{3} e^{3 x^{2}} d x

x \to 1 + lim (\frac{x ^{2} - 4}{x - 1})

\int \frac{20000}{( x + 100 ) ^{3}} d x

\int 8 x e^{x^{2}} d x