What is the derivative of (x^2/(x^4+16)) ?

The derivative of (x^2/(x^4+16)) is (-2x^5+32x)/((x^4+16)^2)

What is the first derivative of (x^2/(x^4+16)) ?

The first derivative of (x^2/(x^4+16)) is (-2x^5+32x)/((x^4+16)^2)

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\frac{d}{d x} (\frac{x ^{2}}{x ^{4} + 16})

\frac{- 2 x ^{5} + 32 x}{( x ^{4} + 16 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2}}{x ^{4} + 16})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{2} ) ( x ^{4} + 16 ) - \frac{d}{d x} ( x ^{4} + 16 ) x ^{2}}{( x ^{4} + 16 ) ^{2}}

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

\frac{d}{d x} (x^{4} + 16) = 4 x^{3}

= \frac{2 x ( x ^{4} + 16 ) - 4 x ^{3} x ^{2}}{( x ^{4} + 16 ) ^{2}}

Simplify

\frac{2 x ( x ^{4} + 16 ) - 4 x ^{3} x ^{2}}{( x ^{4} + 16 ) ^{2}} : \frac{- 2 x ^{5} + 32 x}{( x ^{4} + 16 ) ^{2}}

= \frac{- 2 x ^{5} + 32 x}{( x ^{4} + 16 ) ^{2}}

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

f (x) = e^{6}

\frac{\partial}{\partial x} (ln ((x y)^{z}))

\frac{\partial}{\partial x} (x^{2} + w^{2})

x \to 0 - lim (ln (sin (x)))

What is the derivative of (x^2/(x^4+16)) ?
The derivative of (x^2/(x^4+16)) is (-2x^5+32x)/((x^4+16)^2)
What is the first derivative of (x^2/(x^4+16)) ?
The first derivative of (x^2/(x^4+16)) is (-2x^5+32x)/((x^4+16)^2)