d dx [ e^ ax (bx + c)]= - Vidyadip

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
${e^{ax}}[\cos (bx + c) - \sin (bx + c)]$
D
None of these

✓

Answer

Correct option: A.

${e^{ax}}[a\cos (bx + c) - b\sin (bx + c)]$

a
(a) $\frac{d}{{dx}}[{e^{ax}}\cos (bx + c)]$=$\frac{{dx}}{{dt}} = - 2\sin t + 2\sin 2t$

=${e^{ax}}[a\cos (bx + c) - b\sin (bx + c)]$.

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