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

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

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

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

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derivative of

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

Solution steps

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{4} + 4 x^{3}) = 12 x^{3} + 12 x^{2}

= \frac{1}{3 ( 3 x ^{4} + 4 x ^{3} ) ^{\frac{2}{3}}} (12 x^{3} + 12 x^{2})

Simplify

\frac{1}{3 ( 3 x ^{4} + 4 x ^{3} ) ^{\frac{2}{3}}} (12 x^{3} + 12 x^{2}) : \frac{4 x ^{2} ( x + 1 )}{( x ^{3} ) ^{\frac{2}{3}} ( 3 x + 4 ) ^{\frac{2}{3}}}

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

\int cos^{2} (t) - sin^{2} (t) d t

\frac{\partial}{\partial x} (sin (x^{1} y^{3}))

(- 2 cos (2 x) - sin (3 x))^{2}

x \to \frac{1}{2} lim (arcsin (x))

f (x) = x^{2} + x - 3

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