Derivatives involving trigonometry

**Melnarie** · January 16th 2007, 05:45 PM

"Given y=2sin2t and x=5cos4t, determine dy/dx."

I know I have to use the chain rule here by:

dy/dx = dy/dt x dt/dx but I'm not sure how to isolate dy/dt and dt/dx.

Please help.

**topsquark** · January 16th 2007, 05:55 PM

Originally Posted by Melnarie

"Given y=2sin2t and x=5cos4t, determine dy/dx."

I know I have to use the chain rule here by:

dy/dx = dy/dt x dt/dx but I'm not sure how to isolate dy/dt and dt/dx.

Please help.

There's a slightly easier way.
$y = 2 sin(2t)$ ==> $\frac{dy}{dt} = 2 cos(2t) \cdot 2 = 4 cos(2t)$

$x = 5 cos(4t)$ ==> $\frac{dx}{dt} = 5 \cdot -sin(4t) \cdot 4 = -20 sin(4t)$

Thus
$\frac{dy}{dx} = \frac{ \left ( \frac{dy}{dt} \right ) }{ \left ( \frac{dx}{dt} \right ) }$

$\frac{dy}{dx} = \frac{4 cos(2t)}{-20 sin(4t)} = - \frac{cos(2t)}{5sin(4t)}$

-Dan

**Melnarie** · January 16th 2007, 06:08 PM

oooh! thanks so much!!

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