How I get the nth derivative of the function $y = e^x x^2$ I'm totally confused. Please anyone help me.
$y' = e^x x^2 + 2 e^xx$
$y''= e^x x^2 + 4 e^xx + 2e^x$
$y'''= e^x x^2 + 6 e^xx + 6 e^x$
next $y''''$ but I failed to get any pattern.
 A: $y^{(1)}=\left(x^2+2x\right)e^x$
$y^{(2)}=\left(x^2+4x+2\right)e^x$
$y^{(3)}=\left(x^2+6x+6\right)e^x$
$y^{(4)}=\left(x^2+8x+12\right)e^x$
$y^{(5)}=\left(x^2+10x+20\right)e^x$
where $y^{n}$ is the nth derivative of the function
You see a pattern here where the $x^2$ is constant, the $x$ term increases by 2 every derivative, and the constant term is $(n-0.5)^2-0.25$.
With this pattern we can see that the nth derivative is $y^{n}=\left(x^2+2nx+((n-0.5)^2-0.25)\right)\cdot e^x$
A: For the nth derivative of a function which can be written as product of 2 functions, we can use the formula
$(fg)^n= \sum\limits_{k=0}^{n} \binom {n} {k} f^kg^{n-k}$
And it is quite easy to apply it as any nth derivative of $e^x$ is $e^x$ and differentiation of $x^2$ beyond 2nd derivative is just $0 $
So $(e^xx^2)^n=e^x(\binom {n} {0} x^2 +\binom {n} {1} 2x+ \binom {n} {2} 2)= e^{x}(x^{2}+2nx+n(n-1))
$
A: $$y=e^xx^2$$
$$y'=e^xx^2+2e^xx=y+2xe^x$$
This leaves the result $2xe^x=y'-y$
$$y''=y'+2xe^x+2e^x=y'+(y'-y)+2e^x=2y'-y+2e^x$$
This leaves the result $2e^x=y''+y-2y'$
$$y'''=2y''-y'+2e^x=2y''-y+y''+y-2y'=3y''-2y'$$
The nth derivative for $n\geq4$ leaves:
$$y^{(n)}=3y^{((n-1))}-2y^{((n-2))}$$
A: We need to find an expression to $\frac{d^{n}y}{dx^{n}}$.
Observe that you can use $\underline{\text{product rule}}$ and $\underline{\text{chain rule}}$ in each calculation of the derivative to $y$. Indeed
$$\frac{dy}{dx}=e^{x}(x^{2}+2x+(1-1)(1))$$
$$\frac{d^{2}y}{dx^{2}}=e^{x}(x^{2}+2(2)x+(2-1)(2))$$
$$\frac{d^{3}y}{dx^{3}}=e^{x}(x^{2}+2(3)x+(3-1)(3))$$
$$\frac{d^{4}y}{dx^{4}}=e^{x}(x^{2}+2(4)x+(4-1)(4))$$
$$\frac{d^{5}y}{dx^{5}}=e^{x}(x^{2}+2(5)x+(5-1)(5))$$
so, by an inductive argument, we can see that
$$\boxed{\frac{d^{n}y}{dx^{n}}=e^{x}(x^{2}+2nx+(n-1)(n)), \quad n \in \mathbb{N}}$$
A: Consider the function:
$$f(\lambda,x) = \exp(\lambda x)$$
Then the function you want to differentiate n times w.r.t. x is $ \frac{\partial^2 f}{\partial\lambda^2}$ at $\lambda = 1$. So, we want to evaluate:
$$ \frac{\partial^n}{\partial x^n}\frac{\partial^2 f}{\partial\lambda^2}$$
We can then interchange the order of differentiation to write this as:
$$ \frac{\partial^2}{\partial \lambda^2}\frac{\partial^n f}{\partial x^n}$$
We have:
$$\frac{\partial^n f}{\partial x^n} = \lambda^n \exp(\lambda x)$$
differentiating twice w.r.t. $\lambda$ gives:
$$\left[n(n-1)\lambda^{n-2} + 2 n\lambda^{n-1} x + \lambda^n x^2\right]\exp(\lambda x)$$
Putting $\lambda = 1$ yields the result:
$$\left[n(n-1)+ 2 n x + x^2\right]\exp(x)$$
A: We can also use Taylor series to do it. I don't say that it's the fastest and easiest method, but it's a possibility.
Let's define $f$ as
$$f(x):=x^2e^x$$
and let's fix a constant $a \in \mathbb{R}$. With this, we can define a new function $g$:
$$g(x):=f(x+a)=(x+a)^2e^{x+a}$$
With the chain rule, we can easily see that
$$g^{(n)}(0)=f^{(n)}(a)$$
So if we can find the Maclaurin series of $g$, we are done. But it's easy, because:
\begin{align}
g(x)
&=(x+a)^2e^{x+a}\\
&=(x^2+2ax+a^2)e^ae^x \\
&=(x^2+2ax+a^2)e^a\sum_{n=0}^{+\infty}\frac{x^n}{n!} \\
&=e^a \sum_{n=0}^{+\infty}\frac{x^{n+2}}{n!}+2ae^a \sum_{n=0}^{+\infty}\frac{x^{n+1}}{n!}+a^2e^a \sum_{n=0}^{+\infty}\frac{x^{n}}{n!}\\
&=e^a \sum_{n=2}^{+\infty}\frac{x^{n}}{(n-2)!}+2ae^a \sum_{n=1}^{+\infty}\frac{x^{n}}{(n-1)!}+a^2e^a \sum_{n=0}^{+\infty}\frac{x^{n}}{n!}\\
&=e^a \sum_{n=2}^{+\infty}x^n\frac{n(n-1)}{n!}+2ae^a \sum_{n=1}^{+\infty}x^n\frac{n}{n!}+a^2e^a \sum_{n=0}^{+\infty}\frac{x^{n}}{n!}\\
&=e^a \sum_{n=0}^{+\infty}x^n\frac{n(n-1)}{n!}+2ae^a \sum_{n=0}^{+\infty}x^n\frac{n}{n!}+a^2e^a \sum_{n=0}^{+\infty}\frac{x^{n}}{n!}\\
&=\sum_{n=0}^{+\infty}x^n\Bigg(e^a \frac{n(n-1)}{n!}+2ae^a\frac{n}{n!}+a^2e^a\frac{1}{n!} \Bigg)\\
&=\sum_{n=0}^{+\infty}\frac{x^n}{n!}\Bigg(e^a n(n-1)+2ae^an+a^2e^a \Bigg)\\
\end{align}
Which implies that
$$g^{(n)}(0)=e^a n(n-1)+2ae^an+a^2e^a=f^{(n)}(a)$$
