Find the derivative of y - t 2-2- t 2-2-A- 8 t - t 2-22B- 6 t-t2-224 tC- 4 t-t2-22D- 5 t - t 2-22

The correct option is A 8 t(t^2+2)^2We apply the Quotient Rule with u=t^2 2 and v=t^2+2 d (u/v)dt=(v(du/dt) u(dv/dt))v^2 dydt=((t^2+2).2t (t^2 2).2t)(t^2+2)^2 ...

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Byju's Answer

Standard XII

Physics

Basic Differentiation Rule

Find the deri...

Question

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

\frac{8 t}{(t^{2} + 2)^{2}}

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\frac{6 t}{(t^{2} + 2)^{2}}

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

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\frac{5 t}{(t^{2} + 2)^{2}}

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Solution

The correct option is A $\frac{8 t}{(t^{2} + 2)^{2}}$
We apply the Quotient Rule with $u = t^{2} - 2$ and $v = t^{2} + 2$
$\frac{d (\frac{u}{v})}{d t} = \frac{(v (\frac{d u}{d t}) - u (\frac{d v}{d t}))}{v^{2}}$
$\frac{d y}{d t} = \frac{((t^{2} + 2) .2 t - (t^{2} - 2) .2 t)}{(t^{2} + 2)^{2}}$
$\frac{d y}{d t} = \frac{(2 t^{3} + 4 t - 2 t^{3} + 4 t)}{(t^{2} + 2)^{2}}$
$\frac{d y}{d t} = \frac{8 t}{(t^{2} + 2)^{2}}$

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Similar questions

Q. Find the derivative of

y = \frac{t^{2} - 1}{t^{2} + 1}

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

Q. If

x = \sqrt{\frac{1 - t^{2}}{1 + t^{2}}}

and

y = \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}},

then the value of

\frac{d^{2} y}{d x^{2}}

t = 0

is given by

Q. If

x = \frac{2 t}{1 + t^{2}}

y = \frac{1 - t^{2}}{1 + t^{2}}

, then

\frac{d y}{d x}

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Q. If x = t², y = t³, then

\frac{d^{2} y}{d x^{2}} =

(a) 3/2
(b) 3/4t
(c) 3/2t
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Find the derivative of y=(t2−2)(t2+2).

The correct option is A 8t(t2+2)2We apply the Quotient Rule with u=t2−2 and v=t2+2 d(uv)dt=(v(dudt)−u(dvdt))v2 dydt=((t2+2).2t−(t2−2).2t)(t2+2)2 dydt=(2t3+4t−2t3+4t)(t2+2)2 dydt=8t(t2+2)2

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