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

The derivative of ((4x)/(x^2+1))^{1/4} is (-x^2+1)/(2*2^{1/2)x^{3/4}(x^2+1)^{5/4}}

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

The first derivative of ((4x)/(x^2+1))^{1/4} is (-x^2+1)/(2*2^{1/2)x^{3/4}(x^2+1)^{5/4}}

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = 4 x + 2

\int \frac{( y )}{( 1 + y ^{(2)} )} d y

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

(5 t + 2 y + 6) d t - d y = 0

\frac{d}{d x} (\frac{d y}{d x}) + 1 = 0

What is the derivative of ((4x)/(x^2+1))^{1/4} ?
The derivative of ((4x)/(x^2+1))^{1/4} is (-x^2+1)/(2*2^{1/2)x^{3/4}(x^2+1)^{5/4}}
What is the first derivative of ((4x)/(x^2+1))^{1/4} ?
The first derivative of ((4x)/(x^2+1))^{1/4} is (-x^2+1)/(2*2^{1/2)x^{3/4}(x^2+1)^{5/4}}