Derivative of $x^2e^{-x(y+c)}$

Question

What's the derivative of this function on x

$x^2e^{-x(y+c)}$

And why? I'm not sure about the result I have on the book

y and c are both constants

Answer 1

In derivation, like in many other fields in life, one often has to go slowly to ensure the correct answer was achieved.

In this case, we follow closely the product rule: $\frac{\operatorname d}{\operatorname dx}(f\cdot g) = \frac{\operatorname df}{\operatorname dx}\cdot g + f\cdot\frac{\operatorname dg}{\operatorname dx}$

We also use the following identities:

1. $\frac{\operatorname dx^k}{\operatorname dx} = kx^{k-1}$
2. $\frac{\operatorname de^{f(x)}}{\operatorname dx} = \frac{\operatorname df}{\operatorname dx}\cdot e^{f(x)}$

Therefore we have:

$$\begin{align} \frac{\operatorname d}{\operatorname dx}(x^2e^{-x(y+c)}) &= \frac{\operatorname d}{\operatorname dx}(x^2)\cdot e^{-x(y+c)} + x^2\cdot\frac{\operatorname d}{\operatorname dx}(e^{-x(y+c)}) \\ &=2xe^{-x(y+c)} + x^2(-y-c)e^{-x(y+c)} \\ &=e^{-x(y+c)}(2x-(y+c)x^2) \\ &=xe^{-x(y+c)}(2-(y+c)x) \end{align}$$

Answer 2

Let $z= x^{2}e^{-x (y+c)}$. If $y$ is a function of $x$, then your derivative becomes $$\frac{\rm dz}{\rm dx} = x^{2} e^{-x(y+c)} \times \frac{\rm d}{\rm dx} \bigl( -xy\bigr) + e^{-x(y+c)}\cdot 2x $$

Now $$ \frac{\rm d}{\rm dx}(xy) = x \frac{\rm dy}{ \rm dx } + y$$

Now if $y$ is not a function of $x$, then consider it as a constant and use the chain rule to differentiate.