Question

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Answer 1

Question of Differentiation, Using Chain rule

Given, y = cos ( 5 - 3t)

Now,

$\frac{dy}{dt} = \frac{d}{dt} (cos5 - 3t)$

By chain rule,

$= \frac{d(cos5 - 3t)}{d(5 - 3t)} \times \frac{d(5 - 3t)}{t} \\ \\ = - sin(5 - 3t) \times( 0 - 3 \times 1) \\ \\ = - 3 \times - sin(5 - 3t) \\ \\ = 3 \: sin(5 - 3t)$

Therefore,

If y = cos ( 5 - 3t), dy/dt = 3 sin(5-3t)

Formulae used,

• dy/dt = dy/dx * dx/dt ( Chain rule)
• d/dx ( cosx) = - sinx
• d/dx ( u + v) = d/dx(u) +d/dx(v)
• d/dx( ky) = k d/dx(y), k is a constant.
• d/dx(x^n) = nx^(n-1)

Answer 2

Solution :-

We have

$y = cos(5-3t)$

Now

$\dfrac{dy}{dt}$

$= \dfrac{d (cos(5-3t))}{dt}$

Now as it cannot be directly differentiated so we will apply chain rule.

$\dfrac{d(f(g))}{dk} = \dfrac{d(f(g))}{dg} \times \dfrac{d(g)}{dk}$

or if there are more terms then the chain will continue.

Now back to the question :-

$\implies \dfrac{d (cos(5-3t))}{dt} = \dfrac{d(cos(5-3t))}{d(5-3t)} \times \dfrac{d(5-3t)}{dt}$

Now as

$\dfrac{d(cos(x))}{dx} = -sin(x)$

$\dfrac{d(cx + k)}{dx} = c$

Then

$= \dfrac{d(cos(5-3t))}{d(5-3t)} \times \dfrac{d(5-3t)}{dt}$

$= -sin(5-3t) \times (-3)$

$= 3 sin(5-3t)$