What is the derivative of \sqrt[4]{2x^2-4x+7} ?

The derivative of \sqrt[4]{2x^2-4x+7} is (x-1)/((2x^2-4x+7)^{3/4)}

What is the first derivative of \sqrt[4]{2x^2-4x+7} ?

The first derivative of \sqrt[4]{2x^2-4x+7} is (x-1)/((2x^2-4x+7)^{3/4)}

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\frac{x - 1}{( 2 x ^{2} - 4 x + 7 ) ^{\frac{3}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( 2 x ^{2} - 4 x + 7 ) ^{\frac{3}{4}}} \frac{d}{d x} (2 x^{2} - 4 x + 7)

= \frac{1}{4 ( 2 x ^{2} - 4 x + 7 ) ^{\frac{3}{4}}} \frac{d}{d x} (2 x^{2} - 4 x + 7)

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

= \frac{1}{4 ( 2 x ^{2} - 4 x + 7 ) ^{\frac{3}{4}}} (4 x - 4)

Simplify

\frac{1}{4 ( 2 x ^{2} - 4 x + 7 ) ^{\frac{3}{4}}} (4 x - 4) : \frac{x - 1}{( 2 x ^{2} - 4 x + 7 ) ^{\frac{3}{4}}}

= \frac{x - 1}{( 2 x ^{2} - 4 x + 7 ) ^{\frac{3}{4}}}

\int ((1 - (\frac{x ^{2}}{9})))^{\frac{1}{2}} d x

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

\int (x + 2) (x - 5) d x

\frac{d}{d x} (e^{\frac{x ^{4}}{10}})

\frac{d y}{d x} = x^{2} + y - 1

What is the derivative of \sqrt[4]{2x^2-4x+7} ?
The derivative of \sqrt[4]{2x^2-4x+7} is (x-1)/((2x^2-4x+7)^{3/4)}
What is the first derivative of \sqrt[4]{2x^2-4x+7} ?
The first derivative of \sqrt[4]{2x^2-4x+7} is (x-1)/((2x^2-4x+7)^{3/4)}