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Popular >

derivative of \sqrt[5]{x^2-x+7}

Solution

Solution

\frac{2 x - 1}{5 ( x ^{2} - x + 7 ) ^{\frac{4}{5}}}

Solution steps

Apply the chain rule:

\frac{1}{5 ( x ^{2} - x + 7 ) ^{\frac{4}{5}}} \frac{d}{d x} (x^{2} - x + 7)

= \frac{1}{5 ( x ^{2} - x + 7 ) ^{\frac{4}{5}}} \frac{d}{d x} (x^{2} - x + 7)

\frac{d}{d x} (x^{2} - x + 7) = 2 x - 1

= \frac{1}{5 ( x ^{2} - x + 7 ) ^{\frac{4}{5}}} (2 x - 1)

Simplify

\frac{1}{5 ( x ^{2} - x + 7 ) ^{\frac{4}{5}}} (2 x - 1) : \frac{2 x - 1}{5 ( x ^{2} - x + 7 ) ^{\frac{4}{5}}}

= \frac{2 x - 1}{5 ( x ^{2} - x + 7 ) ^{\frac{4}{5}}}

Popular Examples

(\partial)/(\partial z)(2y^3z)

\frac{\partial}{\partial z} (2 y^{3} z)

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

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

integral of x^5(cos(x^3))

\int x^{5} (cos (x^{3})) d x

d/(dz)(ln(sqrt((a^2-z^2)/(a^2+z^2))))

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Frequently Asked Questions (FAQ)

What is the derivative of \sqrt[5]{x^2-x+7} ?
The derivative of \sqrt[5]{x^2-x+7} is (2x-1)/(5(x^2-x+7)^{4/5)}
What is the first derivative of \sqrt[5]{x^2-x+7} ?
The first derivative of \sqrt[5]{x^2-x+7} is (2x-1)/(5(x^2-x+7)^{4/5)}