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

The derivative of \sqrt[3]{(4x^2+3x-2^2}) is (2(8x+3))/(3(4x^2+3x-2)^{1/3)}

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

The first derivative of \sqrt[3]{(4x^2+3x-2^2}) is (2(8x+3))/(3(4x^2+3x-2)^{1/3)}

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\frac{2 ( 8 x + 3 )}{3 ( 4 x ^{2} + 3 x - 2 ) ^{\frac{1}{3}}}

Solution steps

Apply the chain rule:

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

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

\frac{d}{d x} ((4 x^{2} + 3 x - 2)^{2}) = 2 (4 x^{2} + 3 x - 2) (8 x + 3)

= \frac{1}{3 ( ( 4 x ^{2} + 3 x - 2 ) ^{2} ) ^{\frac{2}{3}}} \cdot 2 (4 x^{2} + 3 x - 2) (8 x + 3)

Simplify

\frac{1}{3 ( ( 4 x ^{2} + 3 x - 2 ) ^{2} ) ^{\frac{2}{3}}} \cdot 2 (4 x^{2} + 3 x - 2) (8 x + 3) : \frac{2 ( 8 x + 3 )}{3 ( 4 x ^{2} + 3 x - 2 ) ^{\frac{1}{3}}}

= \frac{2 ( 8 x + 3 )}{3 ( 4 x ^{2} + 3 x - 2 ) ^{\frac{1}{3}}}

What is the derivative of \sqrt[3]{(4x^2+3x-2^2}) ?
The derivative of \sqrt[3]{(4x^2+3x-2^2}) is (2(8x+3))/(3(4x^2+3x-2)^{1/3)}
What is the first derivative of \sqrt[3]{(4x^2+3x-2^2}) ?
The first derivative of \sqrt[3]{(4x^2+3x-2^2}) is (2(8x+3))/(3(4x^2+3x-2)^{1/3)}