How do you differentiate #y=tan[ln(1-3x)]#?
Apply The Chain Rule:
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To differentiate ( y = \tan[\ln(1-3x)] ), use the chain rule. The derivative is:
[ \frac{dy}{dx} = \sec^2[\ln(1-3x)] \cdot \frac{d}{dx}[\ln(1-3x)] ]
[ = \sec^2[\ln(1-3x)] \cdot \frac{1}{1-3x} \cdot \frac{d}{dx}[-3x] ]
[ = \sec^2[\ln(1-3x)] \cdot \frac{1}{1-3x} \cdot (-3) ]
[ = -3\sec^2[\ln(1-3x)] \cdot \frac{1}{1-3x} ]
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When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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