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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{sin ( x )}{1 + x ^{2}} d x

(x^{12})^{5}

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

\frac{\partial}{\partial x} (3 x y e^{- x^{2} y})

n = 2 \sum \infty \frac{1}{n - 1} - \frac{1}{n}

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