Is this correct? $ {d \over dy} (1+xy)^y = (1+xy)^y \cdot (1+x \cdot \ln(1+xy))$ I know the formula $ {d \over dx} x^x = x^x \cdot( 1+ \ln x ) $, but is below evaluation correct?
$ {d \over dy} (1+xy)^y = (1+xy)^y \cdot (1+x \cdot \ln(1+xy))$
 A: Hint
For this kind of expressions, logarithmic differentiation makes life easier $$A=(1+xy)^y$$ $$\log(A)=y\log(1+xy)$$ Now, differentiate and use the product rule for the rhs.$$\frac{A'}A=\log(1+xy)+y \frac x {1+xy}=\log(1+xy)+ \frac {xy} {1+xy}$$ Now, multiply the lhs and rhs by $A$ and you can (a little) simplify.
A: One has
$$\frac{\mathrm{d}}{\mathrm{d} y}\left(1+xy\right)^{y}=\frac{\mathrm{d}}{\mathrm{d} y}\mathrm{e}^{y\ln\left(1+xy\right)}=\mathrm{e}^{y\ln\left(1+xy\right)}\frac{\mathrm{d}}{\mathrm{d} y}\left[y\ln\left(1+xy\right)\right]=\left(1+xy\right)^{y}\frac{\mathrm{d}}{\mathrm{d} y}\left[y\ln\left(1+xy\right)\right]$$
and
$$\frac{\mathrm{d}}{\mathrm{d} y}\left[y\ln\left(1+xy\right)\right]=\ln\left(1+xy\right)+\frac{xy}{1+xy}.$$
Hence, one formally has

$$\frac{\mathrm{d}}{\mathrm{d} y}\left(1+xy\right)^{y}=\left(1+xy\right)^{y}\ln\left(1+xy\right)+xy\left(1+xy\right)^{y-1}.$$

A: In general, $$\frac{d}{dx}f(x)^{g(x)}=f(x)^{g(x)}\frac{d}{dx}\left(g(x)\ln f(x)\right).$$ Therefore we have $$\frac{d}{dy}(1+xy)^y=(1+xy)^y\frac{d}{dy}\left(y\ln(1+xy)\right)\\ \frac{d}{dy}(1+xy)^y=\color\red{(1+xy)^y\left(\ln(1+xy)+\frac{xy}{1+xy}\right)}. $$
A: $y=f(x)^{g(x)}$ then $y^{'}=g(x)^{'}f(x)^{g(x)}log(f(x))+g(x)f(x)^{g(x)-1}f(x)^{'}$ 
See:
$y=f(x)^{g(x)}$; $log(y)=g(x)log(f(x))$; $\frac{y^{'}}{y}=g(x)^{'}log{f(x)}+g(x)\frac{f(x)^{'}}{f(x)}$; $y^{'}=g(x)^{'}log{f(x)}y+g(x)\frac{f(x)^{'}}{f(x)}y$; $y^{'}=g(x)^{'}f(x)^{g(x)}log(f(x))+g(x)\frac{f(x)^{'}}{f(x)}f(x)^{g(x)}$ 
