What is the (sin(x^3))^' ?

The (sin(x^3))^' is cos(x^3)*3x^2

What is the first (sin(x^3))^' ?

The first (sin(x^3))^' is cos(x^3)*3x^2

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(sin (x^{3}))^{^{'}}

cos (x^{3}) \cdot 3 x^{2}

Solution steps

(sin (x^{3}))^{'}

Apply the chain rule:

cos (x^{3}) (x^{3})^{'}

= cos (x^{3}) (x^{3})^{'}

(x^{3})^{'} = 3 x^{2}

= cos (x^{3}) \cdot 3 x^{2}

x \to - 1 lim (ln (1 + x^{3}))

f (x) = x^{3} - 7 x + 6, at x = - 1

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

\int \frac{x - 4}{( x + 2 ) ^{2} + 16} d x

x \to 0 lim (\frac{e ^{x^{3}} - 1}{x})