In one line
\(D^n\left[e^{ax}\sin(bx+c)\right]=r^ne^{ax}\sin(bx+c+n\phi)\) with \(r=\sqrt{a^2+b^2}\), \(\phi=\tan^{-1}\dfrac ba\); each derivative multiplies by \(r\) and adds the angle \(\phi\).
Explained simply
This function is a sine wave whose size grows (or shrinks) exponentially, like the vibration of a car's suspension or a guitar string dying down. When you differentiate it, two terms appear: one from \(e^{ax}\) and one from the sine. They can be combined into a single sine with a bigger amplitude \(r\) and a shifted angle \(\phi\). Repeating \(n\) times gives \(r^n\) and \(n\phi\).
Technically, \(y_1=e^{ax}[a\sin(bx+c)+b\cos(bx+c)]\). Put \(a=r\cos\phi\), \(b=r\sin\phi\), so \(r=\sqrt{a^2+b^2}\), \(\phi=\tan^{-1}(b/a)\). Then \(y_1=re^{ax}\sin(bx+c+\phi)\). This has the same shape as \(y\), so the step repeats.
So the whole job reduces to two small calculations, \(r\) and \(\phi\), followed by writing down the pattern. This is much quicker and safer than differentiating the product again and again with the product rule, where the number of terms keeps doubling.
How it works
- Identify \(a\) (coefficient in the exponent) and \(b\) (coefficient of \(x\) in the sine).
- Compute \(r=\sqrt{a^2+b^2}\) and \(\phi=\tan^{-1}(b/a)\).
- Write \(y_n=r^ne^{ax}\sin(bx+c+n\phi)\) (or cos).
- For products like \(e^{x}\sin x\cos x\) or \(e^{x}\sin^2x\), first convert to sums: \(e^x\sin x\cos x=\frac12e^x\sin2x\).
Key formula
\(D^n\left[e^{ax}\sin(bx+c)\right]=(a^2+b^2)^{n/2}e^{ax}\sin\left(bx+c+n\tan^{-1}\dfrac ba\right)\)
\(D^n\left[e^{ax}\cos(bx+c)\right]=(a^2+b^2)^{n/2}e^{ax}\cos\left(bx+c+n\tan^{-1}\dfrac ba\right)\)
Quick example
\(y=e^x\sin x\): \(a=1\), \(b=1\), \(r=\sqrt2\), \(\phi=\pi/4\). So \(y_n=2^{n/2}e^x\sin\left(x+\dfrac{n\pi}4\right)\).
Check \(n=1\): \(\sqrt2e^x\sin(x+\pi/4)=\sqrt2e^x\left(\frac{\sin x+\cos x}{\sqrt2}\right)=e^x(\sin x+\cos x)\), which is the direct derivative. Correct.
\(y=e^{2x}\cos^2x=\frac12e^{2x}+\frac12e^{2x}\cos2x\): \(y_n=2^{n-1}e^{2x}+\frac12(2\sqrt2)^ne^{2x}\cos\left(2x+\frac{n\pi}4\right)\), since \(r=\sqrt{4+4}=2\sqrt2\), \(\phi=\tan^{-1}1=\frac\pi4\).
Common mistakes
- Using \(\phi=\tan^{-1}(a/b)\) instead of \(\tan^{-1}(b/a)\).
- Writing \(r^n\) as \((a^2+b^2)^n\) instead of \((a^2+b^2)^{n/2}\).
- Forgetting that \(e^{ax}\) is unchanged (no \(a^n\) in front separately).
How MU asks it
Asked as 3-mark Q.1 parts ("Find \(y_n\) for \(e^{3x}\sin4x\)": \(r=5\), \(y_n=5^ne^{3x}\sin(4x+n\tan^{-1}\frac43)\)) or 5-mark parts with products such as \(e^{x}\sin x\sin2x\). Quote the standard result, show \(r\) and \(\phi\), convert products to sums first.