What is the derivative of \sqrt[3]{(3x^2+6x-4^2}) ?

The derivative of \sqrt[3]{(3x^2+6x-4^2}) is (4(x+1))/((3x^2+6x-4)^{1/3)}

What is the first derivative of \sqrt[3]{(3x^2+6x-4^2}) ?

The first derivative of \sqrt[3]{(3x^2+6x-4^2}) is (4(x+1))/((3x^2+6x-4)^{1/3)}

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\frac{4 ( x + 1 )}{( 3 x ^{2} + 6 x - 4 ) ^{\frac{1}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( ( 3 x ^{2} + 6 x - 4 ) ^{2} ) ^{\frac{2}{3}}} \frac{d}{d x} ((3 x^{2} + 6 x - 4)^{2})

= \frac{1}{3 ( ( 3 x ^{2} + 6 x - 4 ) ^{2} ) ^{\frac{2}{3}}} \frac{d}{d x} ((3 x^{2} + 6 x - 4)^{2})

\frac{d}{d x} ((3 x^{2} + 6 x - 4)^{2}) = 2 (3 x^{2} + 6 x - 4) (6 x + 6)

= \frac{1}{3 ( ( 3 x ^{2} + 6 x - 4 ) ^{2} ) ^{\frac{2}{3}}} \cdot 2 (3 x^{2} + 6 x - 4) (6 x + 6)

Simplify

\frac{1}{3 ( ( 3 x ^{2} + 6 x - 4 ) ^{2} ) ^{\frac{2}{3}}} \cdot 2 (3 x^{2} + 6 x - 4) (6 x + 6) : \frac{4 ( x + 1 )}{( 3 x ^{2} + 6 x - 4 ) ^{\frac{1}{3}}}

= \frac{4 ( x + 1 )}{( 3 x ^{2} + 6 x - 4 ) ^{\frac{1}{3}}}

\frac{\partial}{\partial x} (x e^{y} + 2 y)

\int 4 x ln (x) d x

\int \frac{2 x ^{3}}{( x ^{4} + 5 )} d x

1.12 x^{3} - 6.3 x^{2} - 11.2 x + 1.2

x^{2} e^{x}

What is the derivative of \sqrt[3]{(3x^2+6x-4^2}) ?
The derivative of \sqrt[3]{(3x^2+6x-4^2}) is (4(x+1))/((3x^2+6x-4)^{1/3)}
What is the first derivative of \sqrt[3]{(3x^2+6x-4^2}) ?
The first derivative of \sqrt[3]{(3x^2+6x-4^2}) is (4(x+1))/((3x^2+6x-4)^{1/3)}