Question

help pls.... using leibnitz's theorm find nth differential coefficient of cos2x.x^2

Answer 1

obtained this step

Answer 2

but the final answer is :

Answer 3

well lets start by breaking it down

Answer 4

Let's see if that second formula is correct. I would say it's not because it doesn't account for the \(2x\) within the cosine. It's more likely the formula for the \(n\)th derivative of \(x^2\cos x\). \[\begin{align*}
f(x)&=x^2\cos2x\\\\
f'(x)&=2x\cos2x-2x^2\sin2x
\end{align*}\]
The given answer says
\[f^{(n)}(x)=x^2\cos\left(x+\frac{n\pi}{2}\right)+2nx\cos\left(x+\frac{(n-1)\pi}{2}\right)\\
\quad\quad\quad\quad+n(n-1)\cos\left(x+\frac{(n-2)\pi}{2}\right)\]
which would imply
\[f'(x)=x^2\cos\left(x+\frac{\pi}{2}\right)+2x\cos(x-\pi)=x^2\sin x-2x\cos x\]
which is not the same as the actual computed derivative.

Answer 5

thank you!!.. i confused with the question. As you said, the second one is the final answer for x^2 cosx and the first one is for x^2cos2x