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

Question

Charles Arocha · Accepted Answer

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

Using the standard results: # d/dx(tanx)=sec^2x # and #d/dxlnx=1/x# Along with the chain rule then differentiating: # y = tan(ln(2x+1)) # gives: # dy/dx = sec^2(ln(2x+1)) * d/dx ln(2x+1) # # \ \ \ \ \ = sec^2(ln(2x+1)) * 1/(2x+1) * d/dx(2x+1) # # \ \ \ \ \ = sec^2(ln(2x+1)) * 1/(2x+1) * 2 # # \ \ \ \ \ = (2sec^2(ln(2x+1)))/(2x+1) #

Isaiah Allen · Answer

undefined 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} $.