What is the derivative of g(t)=sqrt(1/(t^2-2)) ?

The derivative of g(t)=sqrt(1/(t^2-2)) is -t/((t^2-2)^{3/2)}

What is the first derivative of g(t)=sqrt(1/(t^2-2)) ?

The first derivative of g(t)=sqrt(1/(t^2-2)) is -t/((t^2-2)^{3/2)}

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derivative of

g (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}}}

(5 x^{2} - 3 x + 2)^{7}

\int_{1}^{9} (x - 1) d x

\frac{d y}{d t} = \frac{t ^{2}}{y + t ^{3} y}

\frac{\partial}{\partial x} (ln (7 x))

x^{^{''}} + 3 x^{^{'}} = 0, x (0) = 0, x^{^{'}} (0) = - 3

What is the derivative of g(t)=sqrt(1/(t^2-2)) ?
The derivative of g(t)=sqrt(1/(t^2-2)) is -t/((t^2-2)^{3/2)}
What is the first derivative of g(t)=sqrt(1/(t^2-2)) ?
The first derivative of g(t)=sqrt(1/(t^2-2)) is -t/((t^2-2)^{3/2)}