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

Answer 1

Here you have a function (#tan#) of a function (#ln#) of a function (#ax+b#). You can use the Chain Rule where you derive each function leaving the "nested" one as it is and multiply the derivative together. So you get: #tan(x)# derived gives you: #1/cos^2(x)# #ln(x)# derived gives you: #1/x# #ax+b# derived gives you: #a#.

So finally: #y'=1/(cos^2(ln(ax+b)))*1/(ax+b)*a#

This could alternatively be written as

#y'=(asec^2(ln(ax+b)))/(ax+b)#

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Answer 2

Emilia Aguilar

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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Answer from HIX Tutor

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.