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

The derivative of y= 1/((x^2+4)) is -(2x)/((x^2+4)^2)

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

The first derivative of y= 1/((x^2+4)) is -(2x)/((x^2+4)^2)

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derivative of

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{d}{d x} (x^{3} sin (y^{3}))

y = sin (e^{2} x)

x^{x} - 2^{s} in x

- 0.25 x

What is the derivative of y= 1/((x^2+4)) ?
The derivative of y= 1/((x^2+4)) is -(2x)/((x^2+4)^2)
What is the first derivative of y= 1/((x^2+4)) ?
The first derivative of y= 1/((x^2+4)) is -(2x)/((x^2+4)^2)