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Popular >

derivative of \sqrt[4]{-3x^4-2}

Solution

Solution

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

Popular Examples

integral of (14x^2+9x+43)/((x+2)(x^2+5))

\int \frac{14 x ^{2} + 9 x + 43}{( x + 2 ) ( x ^{2} + 5 )} d x

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\int + sin^{2} (6 x) cos^{2} (6 x) d x

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\int [\frac{1}{x ^{2}} + sec^{2} (π x)] d x

(\partial)/(\partial x)((xy)^{1/2})

\frac{\partial}{\partial x} ((x y)^{\frac{1}{2}})

sum from n=8 to infinity of 1/(4^n)

n = 8 \sum \infty \frac{1}{4 ^{n}}

Frequently Asked Questions (FAQ)

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