What is the derivative of \sqrt[3]{8x^4+9x^3} ?

The derivative of \sqrt[3]{8x^4+9x^3} is (32x^3+27x^2)/(3(x^3)^{2/3)(8x+9)^{2/3}}

What is the first derivative of \sqrt[3]{8x^4+9x^3} ?

The first derivative of \sqrt[3]{8x^4+9x^3} is (32x^3+27x^2)/(3(x^3)^{2/3)(8x+9)^{2/3}}

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derivative of

\frac{32 x ^{3} + 27 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 8 x + 9 ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( 8 x ^{4} + 9 x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (8 x^{4} + 9 x^{3})

= \frac{1}{3 ( 8 x ^{4} + 9 x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (8 x^{4} + 9 x^{3})

\frac{d}{d x} (8 x^{4} + 9 x^{3}) = 32 x^{3} + 27 x^{2}

= \frac{1}{3 ( 8 x ^{4} + 9 x ^{3} ) ^{\frac{2}{3}}} (32 x^{3} + 27 x^{2})

Simplify

\frac{1}{3 ( 8 x ^{4} + 9 x ^{3} ) ^{\frac{2}{3}}} (32 x^{3} + 27 x^{2}) : \frac{32 x ^{3} + 27 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 8 x + 9 ) ^{\frac{2}{3}}}

= \frac{32 x ^{3} + 27 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 8 x + 9 ) ^{\frac{2}{3}}}

(c_{2} sin (8 t))^{^{'}}

\int - 3 x^{5} d x

\int (x - 2) (x + 3) d x

f (x) = x^{\frac{1}{10}}

x \to 0 lim ((1 - 4 x)^{\frac{5}{x}})

What is the derivative of \sqrt[3]{8x^4+9x^3} ?
The derivative of \sqrt[3]{8x^4+9x^3} is (32x^3+27x^2)/(3(x^3)^{2/3)(8x+9)^{2/3}}
What is the first derivative of \sqrt[3]{8x^4+9x^3} ?
The first derivative of \sqrt[3]{8x^4+9x^3} is (32x^3+27x^2)/(3(x^3)^{2/3)(8x+9)^{2/3}}