What is the derivative of \sqrt[4]{3x^2-5x+2} ?

The derivative of \sqrt[4]{3x^2-5x+2} is (6x-5)/(4(3x^2-5x+2)^{3/4)}

What is the first derivative of \sqrt[4]{3x^2-5x+2} ?

The first derivative of \sqrt[4]{3x^2-5x+2} is (6x-5)/(4(3x^2-5x+2)^{3/4)}

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\frac{d}{d x} (4 3 x^{2} - 5 x + 2)

\frac{6 x - 5}{4 ( 3 x ^{2} - 5 x + 2 ) ^{\frac{3}{4}}}

Solution steps

\frac{d}{d x} (4 3 x^{2} - 5 x + 2)

Apply the chain rule:

\frac{1}{4 ( 3 x ^{2} - 5 x + 2 ) ^{\frac{3}{4}}} \frac{d}{d x} (3 x^{2} - 5 x + 2)

= \frac{1}{4 ( 3 x ^{2} - 5 x + 2 ) ^{\frac{3}{4}}} \frac{d}{d x} (3 x^{2} - 5 x + 2)

\frac{d}{d x} (3 x^{2} - 5 x + 2) = 6 x - 5

= \frac{1}{4 ( 3 x ^{2} - 5 x + 2 ) ^{\frac{3}{4}}} (6 x - 5)

Simplify

\frac{1}{4 ( 3 x ^{2} - 5 x + 2 ) ^{\frac{3}{4}}} (6 x - 5) : \frac{6 x - 5}{4 ( 3 x ^{2} - 5 x + 2 ) ^{\frac{3}{4}}}

= \frac{6 x - 5}{4 ( 3 x ^{2} - 5 x + 2 ) ^{\frac{3}{4}}}

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- t}, y (0) = 2, y^{^{'}} (0) = - 1

x^{^{'}} = x^{2} + 1

y^{^{''}} + 6 y^{^{'}} - 5 y = 0

\frac{d}{d x} (e^{5 - 2 x})

x \to - \infty lim (- 8 x)

What is the derivative of \sqrt[4]{3x^2-5x+2} ?
The derivative of \sqrt[4]{3x^2-5x+2} is (6x-5)/(4(3x^2-5x+2)^{3/4)}
What is the first derivative of \sqrt[4]{3x^2-5x+2} ?
The first derivative of \sqrt[4]{3x^2-5x+2} is (6x-5)/(4(3x^2-5x+2)^{3/4)}