What is the derivative of f(t)=(t^2-1)/((t^2+t-2)) ?

The derivative of f(t)=(t^2-1)/((t^2+t-2)) is 1/((t+2)^2)

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

The first derivative of f(t)=(t^2-1)/((t^2+t-2)) is 1/((t+2)^2)

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derivative of

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

\frac{1}{( t + 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

5 x - 1

cos^{2} (5 x^{2} + 6 x + 5)

\frac{4 x ^{3}}{3 - x}

3 x^{2} + 5 x - 5

y = 3 x (4 x - 5) 2

What is the derivative of f(t)=(t^2-1)/((t^2+t-2)) ?
The derivative of f(t)=(t^2-1)/((t^2+t-2)) is 1/((t+2)^2)
What is the first derivative of f(t)=(t^2-1)/((t^2+t-2)) ?
The first derivative of f(t)=(t^2-1)/((t^2+t-2)) is 1/((t+2)^2)