$y = -3^{\sin(x)} - 3^x$ or $y = -[3^{\sin(x)} + 3^x]$. The problem I'm facing is that if I take negative sign out of parentheses and then differentiate it, it's easy, but if I try to differentiate it with the negative sign associated with both terms, then, let's take just $-3^{\sin(x)}$, this happens
$$ \ln y = \sin(x)\ln(-3) $$
This term has $\ln(+3)$ if I don't take negative out of parentheses. Are both ways correct? Is there some kind of identity that can change $\ln(-3)$ to $\ln(3)$ cause I'm sure I can't simply take the sign outside.
Since the derivative of $g(x)=-f_1(x)-f_2(x)$ is $$ g'(x)=-f_1'(x)-f_2'(x) $$ (provided both $f_1$ and $f_2$ are differentiable), you just need to compute the derivatives of $$ f_1(x)=3^{\sin x} \qquad\text{and}\qquad f_2(x)=3^x $$ Here you can use the logarithmic derivative: $$ \ln f_1(x)=\sin x\ln 3 $$ so $$ \frac{f_1'(x)}{f_1(x)}=\cos x\ln 3 $$ and finally $$ f_1'(x)=3^{\sin x}\cos x\ln 3 $$ The derivative of $f_2$ is $f_2'(x)=3^x\ln 3$.