What is the derivative of (3+x-2x^2^{1/4}) ?

The derivative of (3+x-2x^2^{1/4}) is (-4x+1)/(4(3+x-2x^2)^{3/4)}

What is the first derivative of (3+x-2x^2^{1/4}) ?

The first derivative of (3+x-2x^2^{1/4}) is (-4x+1)/(4(3+x-2x^2)^{3/4)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{'}} = \frac{( 2 - y ) ^{2}}{2 \cdot 1 + x}

x = 2 y^{2}, x = 4 + y^{2}

y = x^{3}, y = 9 x

\int 2 cos (x) - sin (x) d x

n = 1 \sum \infty cos (\frac{1}{n ^{2}})

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