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Exercise 5.6 · Q3

Q.Find dydx\frac{dy}{dx} in the following: x=sin⁡t,y=cos⁡2tx = \sin t, y = \cos 2t

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✓ Free question

For parametric equations x=sin⁡tx = \sin t, y=cos⁡2ty = \cos 2t, we use dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt} to get dydx=−4sin⁡t\frac{dy}{dx} = -4 \sin t.

Why parametric differentiation works

When xx and yy are both given in terms of a third variable tt, we cannot directly write yy as a function of xx — and we don't need to. The chain rule gives us a clean way:

dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, provided dx/dt≠0dx/dt \neq 0.

Think of it this way: a small change in tt causes a small change in both xx and yy. The ratio of those changes (as dt→0dt \to 0) is exactly the derivative we want.


Step-by-step solution

1. Differentiate xx with respect to tt

x=sin⁡tx = \sin t

dxdt=cos⁡t\frac{dx}{dt} = \cos t

2. Differentiate yy with respect to tt

y=cos⁡2ty = \cos 2t

Using the chain rule: dydt=−sin⁡(2t)⋅2=−2sin⁡2t\frac{dy}{dt} = -\sin(2t) \cdot 2 = -2 \sin 2t

Tip

You can also use the double-angle identity sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2 \sin t \cos t to rewrite −2sin⁡2t=−4sin⁡tcos⁡t-2 \sin 2t = -4 \sin t \cos t. This will simplify nicely later.

3. Apply the parametric derivative formula

dydx=dy/dtdx/dt=−2sin⁡2tcos⁡t\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2 \sin 2t}{\cos t}

4. Simplify using the identity sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2 \sin t \cos t

dydx=−2(2sin⁡tcos⁡t)cos⁡t=−4sin⁡tcos⁡tcos⁡t\frac{dy}{dx} = \frac{-2 (2 \sin t \cos t)}{\cos t} = \frac{-4 \sin t \cos t}{\cos t}

5. Cancel cos⁡t\cos t (provided cos⁡t≠0\cos t \neq 0, i.e., t≠π2+nπt \neq \frac{\pi}{2} + n\pi)

dydx=−4sin⁡t\frac{dy}{dx} = -4 \sin t

Watch out

A common mistake is to forget the chain rule when differentiating cos⁡2t\cos 2t — the derivative is −2sin⁡2t-2 \sin 2t, not −sin⁡2t-\sin 2t. Also, never cancel cos⁡t\cos t without noting where it is zero; those points correspond to vertical tangents where dx/dt=0dx/dt = 0.


✓Final answer

The derivative is dydx=−4sin⁡t\frac{dy}{dx} = -4 \sin t.

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