# How do you differentiate the function #y=tan[ln(ax + b)]#?

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To differentiate the function ( y = \tan[\ln(ax + b)] ), you would use the chain rule. Here's the process:

- Let ( u = \ln(ax + b) ).
- Compute ( \frac{du}{dx} ) using the chain rule: ( \frac{du}{dx} = \frac{1}{ax + b} \cdot a = \frac{a}{ax + b} ).
- Now differentiate ( y = \tan(u) ) with respect to ( u ): ( \frac{dy}{du} = \sec^2(u) ).
- Multiply ( \frac{dy}{du} ) by ( \frac{du}{dx} ) to get ( \frac{dy}{dx} ).

So, ( \frac{dy}{dx} = \frac{a}{ax + b} \sec^2[\ln(ax + b)] ).

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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.

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