# What is the derivative of #y=tan(ln(2x+1))#?

# dy/dx = (2sec^2(ln(2x+1)))/(2x+1) #

Using the standard results:

Along with the chain rule then differentiating:

gives:

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To find the derivative of ( y = \tan(\ln(2x+1)) ), you can use the chain rule.

Let ( u = \ln(2x+1) ), and ( v = \tan(u) ).

Then, applying the chain rule, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} ]

Now, differentiate ( u = \ln(2x+1) ) with respect to ( x ):

[ \frac{du}{dx} = \frac{d}{dx}(\ln(2x+1)) = \frac{1}{2x+1} \cdot \frac{d}{dx}(2x+1) = \frac{2}{2x+1} ]

Now, differentiate ( v = \tan(u) ) with respect to ( u ):

[ \frac{dv}{du} = \frac{d}{du}(\tan(u)) = \sec^2(u) ]

Finally, combine the results:

[ \frac{dy}{dx} = \frac{2 \sec^2(\ln(2x+1))}{2x+1} ]

So, the derivative of ( y = \tan(\ln(2x+1)) ) is ( \frac{2 \sec^2(\ln(2x+1))}{2x+1} ).

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