## Solution

$y^{\prime}=\left(\ln \left(e^{5 x}+\sqrt{e^{10 x}-1}\right)+\arcsin \left(e^{-5 x}\right)\right)^{\prime}=$

$=\frac{1}{e^{5 x}+\sqrt{e^{10 x}-1}} \cdot\left(5 e^{5 x}+\frac{1}{2 \sqrt{e^{10 x}-1}} \cdot 10 e^{10 x}\right)+\frac{1}{\sqrt{1-e^{-10 x}}} \cdot\left(-5 e^{-5 x}\right)=$

$=\frac{1}{e^{5 x}+\sqrt{e^{10 x}-1}} \cdot \frac{5 e^{5 x} \cdot\left(\sqrt{e^{10 x}-1}+1\right)}{\sqrt{e^{10 x}-1}}-\frac{5 e^{-5 x}}{\sqrt{1-e^{-10 x}}}=$

$=\frac{5 e^{5 x}}{\sqrt{e^{10 x}-1}}-\frac{5 e^{-5 x}}{\sqrt{1-e^{-10 x}}}=\frac{5 e^{5 x} \cdot\left(1-e^{-10 x}\right)}{\sqrt{1-e^{-10 x}}}=$

$=5 e^{5 x} \cdot \sqrt{1-e^{-10 x}}$

## Problem Kuznetsov Differentiation 13-18
## Problem Statement

Find the derivative.

$$
y=\ln \left(e^{5 x}+\sqrt{e^{10 x}-1}\right)+\arcsin \left(e^{-5 x}\right)
$$