Find the derivative of y-t2-1-t2-1-

We apply the Quotient Rule with u=t^2 1 and v=t^2+1 d/dt(u/v)=(v(du/dt) u(dv/dt))/v^2 dy/dt=((t^2+1).2t (t^2 1).2t)/(t^2+1)^2 nbsp; =(2t^3+2t 2t^3+2t)/(t^2+1 ...

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Byju's Answer

Standard XII

Physics

Basic Differentiation Rule

Find the deri...

Question

Find the derivative of $y = \frac{(t^{2} - 1)}{(t^{2} + 1)}$ .

\frac{4 t^{2}}{(t + 1)^{2}}

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\frac{4 t}{(t^{2} + 1)^{3}}

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\frac{4 t}{(t + 1)^{2}}

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\frac{4 t}{(t^{2} + 1)^{2}}

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Solution

The correct option is D $\frac{4 t}{(t^{2} + 1)^{2}}$
We apply the Quotient Rule with $u = t^{2} - 1$ and $v = t^{2} + 1$ $\frac{d}{d t} (u / v) = \frac{(v (d u / d t) - u (d v / d t))}{v^{2}}$ $\frac{d y}{d t} = ((t^{2} + 1) .2 t - (t^{2} - 1) .2 t) / (t^{2} + 1)^{2}$ $= (2 t^{3} + 2 t - 2 t^{3} + 2 t) / (t^{2} + 1)^{2} = \frac{4 t}{(t^{2} + 1)^{2}}$

Suggest Corrections

Find the derivative of y=(t2−1)(t2+1).

The correct option is D 4t(t2+1)2We apply the Quotient Rule with u=t2−1 and v=t2+1 ddt(u/v)=(v(du/dt)−u(dv/dt))v2 dydt=((t2+1).2t−(t2−1).2t)/(t2+1)2 =(2t3+2t−2t3+2t)/(t2+1)2=4t(t2+1)2

Find the derivative of $y = \frac{(t^{2} - 1)}{(t^{2} + 1)}$ .