Answer to Question #287650 in Algebra for Pratham Chavan

Question

Find nth derivative of e^ax cos² xsinx

Answer 1

Solution:

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

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

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

Now, we know that

if $y=e^{ax}$

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

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

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