Find y' for y = ln (cos x + e^-2xsin x) + x

Find $y'$ for $y = \ln\bigl(\cos x + e^{-2x}\sin x\bigr) + x$.

Step 1: Differentiate the inner function

Let $u = \cos x + e^{-2x}\sin x$. The second term is a product, needing both the chain and product rules: \[u' = -\sin x + \Bigl(-2e^{-2x}\sin x + e^{-2x}\cos x\Bigr) = -\sin x + e^{-2x}\bigl(\cos x - 2\sin x\bigr).\]

Step 2: Form the derivative

The added $x$ contributes $1$, so writing everything over $u$: $y' = \frac{u'}{u} + 1 = \frac{u'+u}{u}$.

Step 3: Simplify the numerator

Adding $u$ to $u'$, the $-2\sin x$ inside the bracket is softened to $-\sin x$: \[u'+u = \bigl(\cos x - \sin x\bigr) + e^{-2x}\bigl(\cos x - 2\sin x + \sin x\bigr) = \bigl(\cos x-\sin x\bigr) + e^{-2x}\bigl(\cos x-\sin x\bigr).\]

The factor $\cos x - \sin x$ is now common to both parts: $u'+u = \bigl(\cos x - \sin x\bigr)\bigl(1+e^{-2x}\bigr)$.

Step 4: Verification

\[x = 0:\quad u = 1 \;\Longrightarrow\; y' = \frac{(1+1)(1-0)}{1} = 2; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 2.0 \;\;\checkmark\]

\[x = \frac{\pi}{4}:\quad \cos x - \sin x = 0 \;\Longrightarrow\; y' = 0, \text{ a critical point} \;\;\checkmark\]