Answer to Question #287650 in Algebra for Pratham Chavan

Question #287650

Find nth derivative of e^ax cos² xsinx

1
Expert's answer
2022-02-07T12:20:24-0500

Solution:

"y=e^{ax}\\cos^2x \\sin x\n\\\\\\Rightarrow y=e^{ax}(\\dfrac{\\cos2x+1}2) \\sin x\n\\\\\\Rightarrow y=\\frac12e^{ax}\\cos2x\\sin x+\\frac12e^{ax}\\sin x"

Using "\\cos x=\\dfrac{e^{ix}+e^{-ix}}2, \\sin x=\\dfrac{e^{ix}-e^{-ix}}2"

"y=\\frac12e^{ax}(\\dfrac{e^{2ix}+e^{-2ix}}2)(\\dfrac{e^{ix}-e^{-ix}}2)+\\frac12e^{ax}(\\dfrac{e^{ix}-e^{-ix}}2)\n\\\\=\\frac18e^{ax}(e^{3ix}-e^{ix}+e^{-ix}-e^{-3ix})+\\frac14({e^{(a+i)x}-e^{(a-i)x}})\n\\\\=\\frac18(e^{(a+3i)x}-e^{(a+i)x}+e^{(a+i)x}-e^{(a-3i)x})+\\frac14({e^{(a+i)x}-e^{(a-i)x}})\n\\\\=\\frac18({e^{x\\left(a+3i\\right)}-e^{x\\left(a-3i\\right)}})+\\frac{1}{4}\\left(e^{x\\left(a+i\\right)}-e^{x\\left(a-i\\right)}\\right)"

Now, we know that

if "y=e^{ax}"

then, "y_n=a^ne^{ax}" .

So, "y=\\frac18({e^{x\\left(a+3i\\right)}-e^{x\\left(a-3i\\right)}})+\\frac{1}{4}\\left(e^{x\\left(a+i\\right)}-e^{x\\left(a-i\\right)}\\right)"

"y_n=\\frac18({(a+3i)^ne^{x\\left(a+3i\\right)}-(a-3i)^ne^{x\\left(a-3i\\right)}})+\\frac{1}{4}\\left((a+i)^ne^{x\\left(a+i\\right)}-(a-i)^ne^{x\\left(a-i\\right)}\\right)"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS