How do you find the derivative of # tan 2x = cos 3y#?
using the chain rule:
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To find the derivative of the equation ( \tan(2x) = \cos(3y) ), you would use implicit differentiation. Here's how you do it:
 Take the derivative of both sides of the equation with respect to ( x ).
 For ( \tan(2x) ), you would use the chain rule, and for ( \cos(3y) ), you would use the chain rule as well, treating ( y ) as a function of ( x ) due to implicit differentiation.
 Then, solve for ( \frac{{dy}}{{dx}} ), which represents the derivative of ( y ) with respect to ( x ).
The steps would be:

( \frac{{d}}{{dx}} \tan(2x) = \frac{{d}}{{dx}} \cos(3y) )

Using the chain rule, the derivatives of ( \tan(2x) ) and ( \cos(3y) ) would be:
( \sec^2(2x) \cdot 2 = \sin(3y) \cdot \frac{{d}}{{dx}}(3y) )

Simplify and solve for ( \frac{{dy}}{{dx}} ):
( \frac{{d}}{{dx}}(3y) = 3 \frac{{dy}}{{dx}} )
So, ( 2\sec^2(2x) = 3\sin(3y) \frac{{dy}}{{dx}} )
Therefore, ( \frac{{dy}}{{dx}} = \frac{{2\sec^2(2x)}}{{3\sin(3y)}} )
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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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