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Chain Rule of Differentiation

If y = tan5...

Question

If $y = tan (\frac{5}{2} π t + \frac{π}{6})$ then find the value of $\frac{d y}{d t}$ at $t = 0$ .

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Solution

$y = tan (\frac{5}{2} π t + \frac{π}{6})$
$\frac{d y}{d t} = \frac{5 π}{2} {sec}^{2} (\frac{5}{2} π t + \frac{π}{6})$
Now ${\frac{d y}{d t} ∣ ∣ ∣}_{t = 0} = \frac{5 π}{2} {sec}^{2} \frac{π}{6} = \frac{5 π}{2} \times \frac{4}{3} = \frac{10 π}{3}$ .

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Q. Consider the differential equation

\frac{d^{2} y (t)}{d t^{2}} + 2 \frac{d y (t)}{d t} + y (t) = δ (t) w i t h y (t) |_{t = 0} = - 2 a n d \frac{d y}{d t} |_{t = 0} = 0.

. The numerical value of

\frac{d y}{d t} |_{t = 0}

x = a s i n 2 t (1 + c o s 2 t)

y = b c o s 2 t (1 - c o s 2 t),

then find the values of

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x

y

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\frac{d x}{d t} \neq 0

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