Proving completeness of highest-order statistic using Leibnitz' Rule

Suppose that $$X_1,...,X_n$$ are iid with common pdf given by $$f(x;\theta)=2e^{2x}\theta^{-2}I( x I am tasked with finding a complete-sufficient statistic for $$\theta$$, and I have already shown via Neyman-factorization that $$X_{(n)}$$ is sufficient.

To demonstrate completeness, let $$h(x_{(n)})$$ be a real-valued function of $$x_{(n)}$$ such that $$E[h(x_{(n)})]=0$$. We can now express this as the following integral: $$0=\int_{-\infty}^{log(\theta)}h(x_{(n)})2n\theta^{-2n}e^{2nx_{(n)}}dx_{(n)},$$

where the expression being multiplied by $$h$$ is the pdf of the n-th order statistic. Now, this is where I am a little confused. First, note that because we are integrating w.r.t. the n-th order statistic, we can divide both sides by all expressions in the integral which do not contain the n-th order statistic. We now have the following: $$0=\int_{-\infty}^{log(\theta)}h(x_{(n)})e^{2nx_{(n)}}dx_{(n)}.$$

I am now inclined to use Leibnitz' rule, but in the last expression of Leibnitz' rule, where I take the partial w.r.t. $$\theta$$ of everything in the integral, it will be free of $$h(x_{(n)})$$ which is a problem, as I need to show that $$h(x_{(n)})=0$$.

My question is, should I not have eliminated the seemingly ancillary expressions inside the integral? Or am I making some other type of mistake?

The derivative of $$\int_{-\infty}^{\log(\theta)}h(x)e^{2nx}\text{d}x$$ which is the composition of the two functions $$\theta \mapsto \log(\theta)\qquad\text{ and }\qquad \xi \mapsto \int_{-\infty}^{\xi}h(x)e^{2nx}\text{d}x$$ wrt $$\theta$$ is $$\frac{\text{d}}{\text{d}\theta}\int_{-\infty}^{\log(\theta)}h(x)e^{2nx}\text{d}x=\frac{1}{\theta}\,h(\log(\theta))e^{2n\log(\theta)}=\theta^{2n-1}h(\log(\theta))$$ which is uniformly null iff $$h\equiv 0$$. There is no reason for resorting to Leibniz's rule in that case.