What is the derivative of f(x)=\sqrt[3]{10x^4+11x^3} ?

The derivative of f(x)=\sqrt[3]{10x^4+11x^3} is (40x^3+33x^2)/(3(x^3)^{2/3)(10x+11)^{2/3}}

What is the first derivative of f(x)=\sqrt[3]{10x^4+11x^3} ?

The first derivative of f(x)=\sqrt[3]{10x^4+11x^3} is (40x^3+33x^2)/(3(x^3)^{2/3)(10x+11)^{2/3}}

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derivative of

\frac{40 x ^{3} + 33 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 10 x + 11 ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( 10 x ^{4} + 11 x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (10 x^{4} + 11 x^{3})

= \frac{1}{3 ( 10 x ^{4} + 11 x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (10 x^{4} + 11 x^{3})

\frac{d}{d x} (10 x^{4} + 11 x^{3}) = 40 x^{3} + 33 x^{2}

= \frac{1}{3 ( 10 x ^{4} + 11 x ^{3} ) ^{\frac{2}{3}}} (40 x^{3} + 33 x^{2})

Simplify

\frac{1}{3 ( 10 x ^{4} + 11 x ^{3} ) ^{\frac{2}{3}}} (40 x^{3} + 33 x^{2}) : \frac{40 x ^{3} + 33 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 10 x + 11 ) ^{\frac{2}{3}}}

= \frac{40 x ^{3} + 33 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 10 x + 11 ) ^{\frac{2}{3}}}

(2 - 3 x) cos (3 x)

f (t) = (2 t - 1)^{3}

2 y^{^{'}} = y^{2}

\frac{\partial}{\partial x} (y (- sin (x ln (y))) ln (y))

\frac{d}{d t} (- \frac{1}{t ^{2}})

What is the derivative of f(x)=\sqrt[3]{10x^4+11x^3} ?
The derivative of f(x)=\sqrt[3]{10x^4+11x^3} is (40x^3+33x^2)/(3(x^3)^{2/3)(10x+11)^{2/3}}
What is the first derivative of f(x)=\sqrt[3]{10x^4+11x^3} ?
The first derivative of f(x)=\sqrt[3]{10x^4+11x^3} is (40x^3+33x^2)/(3(x^3)^{2/3)(10x+11)^{2/3}}