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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} + 4}) = \frac{6 x}{( x ^{2} + 4 ) ^{2}}

= \frac{1}{2 \frac{x ^{2} + 1}{x ^{2} + 4}} \cdot \frac{6 x}{( x ^{2} + 4 ) ^{2}}

Simplify

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

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

\frac{\partial}{\partial x} (2 x^{4} y + sin (x^{4}))

\int \frac{arcsin ( ln ( x ) )}{x} d x

\frac{1}{cos ( \frac{x}{2} )}

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

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