How do you differentiate #tan 2x = cos 3y#?
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To differentiate ( \tan(2x) = \cos(3y) ), you can use implicit differentiation. The derivative of both sides of the equation with respect to ( x ) is taken.

Differentiate both sides of the equation with respect to ( x ):
[ \frac{d}{dx}[\tan(2x)] = \frac{d}{dx}[\cos(3y)] ]

Use the chain rule and trigonometric derivative rules:
[ \sec^2(2x) \cdot 2 = \sin(3y) \cdot \frac{d}{dx}(3y) ]

Solve for ( \frac{dy}{dx} ) by isolating ( \frac{d}{dx}(3y) ):
[ \frac{d}{dx}(3y) = \frac{2}{\sec^2(2x)} \cdot \sin(3y) ]
[ \frac{d}{dx}(3y) = 2\sin(3y)\sec^2(2x) ]

Finally, use ( \frac{dy}{dx} = \frac{1}{3} \cdot \frac{d}{dx}(3y) ) to find ( \frac{dy}{dx} ):
[ \frac{dy}{dx} = \frac{2}{3}\sin(3y)\sec^2(2x) ]
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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