MCQ
${d \over {dx}}[{e^{ax}}\cos (bx + c)]=$
- ✓${e^{ax}}[a\cos (bx + c) - b\sin (bx + c)]$
- B${e^{ax}}[a\sin (bx + c) - b\cos (bx + c)]$
- C${e^{ax}}[\cos (bx + c) - \sin (bx + c)]$
- DNone of these
✓
Answer
Correct option: A.
${e^{ax}}[a\cos (bx + c) - b\sin (bx + c)]$
a
(a) $\frac{d}{{dx}}[{e^{ax}}\cos (bx + c)]$=$\frac{{dx}}{{dt}} = - 2\sin t + 2\sin 2t$
(a) $\frac{d}{{dx}}[{e^{ax}}\cos (bx + c)]$=$\frac{{dx}}{{dt}} = - 2\sin t + 2\sin 2t$
=${e^{ax}}[a\cos (bx + c) - b\sin (bx + c)]$.
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