Find derivative of tricky logarithmic functions

Question

Find the derivative of $y=(x^{x+1})(x+1)^x$

So this is what I have,

$$\ln y=\ln[(x^{x+1})(x+1)^x]$$

$$= \ln x^{x+1} + \ln(x+1)^x$$

$$\frac{1}{y}y' = (1)(\ln x) + (x+1)\frac{1}{x} + (1)(\ln(x+1)) + (x)\frac{1}{x+1}$$

$$= \ln x + \frac{x+1}{x} + \ln(x+1) + \frac{x}{x+1}$$

$$y' = (x^{x+1})(x+1)^x \left[\ln x + \ln(x+1) + \frac{x+1}{x} + \frac{x}{x+1}\right]$$

Is that the right answer?

Answer 1

Yes that is the right answer..