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