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

The derivative of x/(x^2+t^2) is (-x^2+t^2)/((x^2+t^2)^2)

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

The first derivative of x/(x^2+t^2) is (-x^2+t^2)/((x^2+t^2)^2)

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\frac{d}{d x} (\frac{x}{x ^{2} + t ^{2}})

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

Solution steps

\frac{d}{d x} (\frac{x}{x ^{2} + t ^{2}})

Treat

t

as a constant

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

1 \cdot (x^{2} + t^{2}) - 2 xx = - x^{2} + t^{2}

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

\int_{0}^{3} - x^{2} d x

\int tan (3 x) sec^{\frac{5}{4}} (3 x) d x

y = x, (36, 6)

\frac{d}{d x} (lo g_{13} (x))

f (x) = 8 x^{2} cot (x)

What is the derivative of x/(x^2+t^2) ?
The derivative of x/(x^2+t^2) is (-x^2+t^2)/((x^2+t^2)^2)
What is the first derivative of x/(x^2+t^2) ?
The first derivative of x/(x^2+t^2) is (-x^2+t^2)/((x^2+t^2)^2)