What is the d/(dt)(sqrt(1/(t^2-2))) ?

The d/(dt)(sqrt(1/(t^2-2))) is -t/((t^2-2)^{3/2)}

What is the first d/(dt)(sqrt(1/(t^2-2))) ?

The first d/(dt)(sqrt(1/(t^2-2))) is -t/((t^2-2)^{3/2)}

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\frac{d}{d t} (\frac{1}{t ^{2} - 2})

- \frac{t}{( t ^{2} - 2 ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d t} (\frac{1}{t ^{2} - 2})

Apply the chain rule:

\frac{1}{2 \frac{1}{t ^{2} - 2}} \frac{d}{d t} (\frac{1}{t ^{2} - 2})

= \frac{1}{2 \frac{1}{t ^{2} - 2}} \frac{d}{d t} (\frac{1}{t ^{2} - 2})

\frac{d}{d t} (\frac{1}{t ^{2} - 2}) = - \frac{2 t}{( t ^{2} - 2 ) ^{2}}

= \frac{1}{2 \frac{1}{t ^{2} - 2}} (- \frac{2 t}{( t ^{2} - 2 ) ^{2}})

Simplify

\frac{1}{2 \frac{1}{t ^{2} - 2}} (- \frac{2 t}{( t ^{2} - 2 ) ^{2}}) : - \frac{t}{( t ^{2} - 2 ) ^{\frac{3}{2}}}

= - \frac{t}{( t ^{2} - 2 ) ^{\frac{3}{2}}}

\frac{\partial}{\partial x} (ln (4 + 3 x^{2} + 2 y^{2}))

f (x) = 2 sin^{4} (x)

y^{^{''}} - 6 y^{^{'}} = - 2 x - 5

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

\int (5 x^{4} - \frac{4}{x ^{4}}) d x

What is the d/(dt)(sqrt(1/(t^2-2))) ?
The d/(dt)(sqrt(1/(t^2-2))) is -t/((t^2-2)^{3/2)}
What is the first d/(dt)(sqrt(1/(t^2-2))) ?
The first d/(dt)(sqrt(1/(t^2-2))) is -t/((t^2-2)^{3/2)}