Find y' for y = e^xcos x
Find $y'$ for $y = e^{x}\cos x$.
Explanation of Method
A product of an exponential and a cosine.
Step 1: Identify the factors and apply the product rule
For $y = uv$ recall $y' = u'v + uv'$. Here
\[ u = e^{x},\quad u' = e^{x}, \qquad v = \cos x,\quad v' = -\sin x. \]
Step 2: Substitute and factor
\[ y' = e^{x}\cos x + e^{x}\bigl(-\sin x\bigr) = e^{x}\bigl(\cos x - \sin x\bigr). \]
Step 3: Verification
\[x = 0:\quad y' = e^{0}(1-0) = 1; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]
\[y' = 0 \;\Longleftrightarrow\; \tan x = 1 \;\Longleftrightarrow\; x = \frac{\pi}{4}+k\pi \;\;\checkmark\]
The factor $\cos x - \sin x$ is exactly the derivative found in Problem 47, which is no coincidence: both come from the same pair of standard derivatives.
Result
Final answer:
\[\boxed{\,y' = e^{x}\bigl(\cos x - \sin x\bigr)\,}\]
Remark
Remark. Since $\cos x - \sin x = \sqrt{2}\cos\!\left(x+\tfrac{\pi}{4}\right)$, the derivative can be written $y' = \sqrt{2}\,e^{x}\cos\!\left(x+\tfrac{\pi}{4}\right)$: each differentiation multiplies the amplitude by $\sqrt{2}$ and advances the phase by $\tfrac{\pi}{4}$.