# Take partial derivative of $\sum_{i=1}^n (y_i - {ae^{x_i^2}} -bx_i^3)^2$ with respect to a

I need to find the partial derivative of $\sum_{i=1}^n (y_i - {ae^{x_i^2}} -bx_i^3)^2$ with respect to a and b but I'm having a hard time with the process. Could someone provide a step by step of how to solve this? I have the answer which is $-2\sum_{i=1}^n (y_i - {ae^{x_i^2}} -bx_i^3)e^{x_i^2}$ for a.

My thought process is that with the power rule and assuming that $bx^3,e^{x^2}$ and $y$ are constants, the answer should be $2\sum_{i=1}^n e^{x_i^2}$.

EDIT: To clarify, Here are the parts I'm unsure of.

1. Why is it -2 instead of just 2 after applying the power rule?
2. Why is $(y_i - {ae^{x_i^2}} -bx_i^3)e^{x_i^2}$ maintained instead of just $e^{x_i^2}$?
• The summation is linear so i can apply it term by term. basically permute the derivative and summation. Can you take thqe partials now? – Chinny84 Mar 16 '17 at 21:37
• Sorry I edited my post to better explain my question – Curious Cat Mar 16 '17 at 21:47
• I'm sorry; I misread your question. When when you say the answer "should be" $2\sum e^{x_i^2}$, what precisely do you mean? The claim $$\frac\partial{\partial a}\sum_{i=1}^n(y_i-ae^{x_i^2}-bx_i^3)^2=2\sum_{i=1}^ne^{x_i^2}$$ seems highly suspect. If we consider the trivial case in which $n=1$ and all other parameters are $0$, then we have $$\frac\partial{\partial a}a^2=2$$ – Kajelad Mar 16 '17 at 22:14

You have to use the chain rule on this \begin{eqnarray*} f(u) &=& u^2 \\ g(a,b,x,y) &=& y_i -ae^{x_i^2} - bx^3_i \end{eqnarray*} so that the above function is a summation over terms looking like $f(g(a,b,x,y))$. Since the derivative is a linear operator you can exchange summation and differentiation to obtain $$\frac{\partial}{\partial a} \sum_i (y_i -ae^{x_i^2} - bx^3_i)^2 = \sum_i \frac{\partial}{\partial a} (y_i -ae^{x_i^2} - bx^3_i)^2$$ The chain rule on the individual pieces states $$\frac{\partial}{\partial a} f(g(a,b,x,y)) = \frac{d f}{d g} \frac{\partial g}{\partial a} =2 (y_i -ae^{x_i^2} - bx^3_i)(-e^{x^2_i})$$ We can then replace the summation to obtain $$\sum_i -2 (y_i -ae^{x_i^2} - bx^3_i)e^{x^2_i} = -2\sum_i(y_i -ae^{x_i^2} - bx^3_i)e^{x^2_i}$$

• Thanks for the answer. My friend explained this to me right before you posted but I understand this now. I haven't done derivatives in a very long time. – Curious Cat Mar 16 '17 at 22:28

Okay, so you need to think about the chain rule and what its implications are. Per your post you have: $\sum_{i=1}^n (y_i - {ae^{x_i^2}} -bx_i^3)^2$ which you are trying to partially differentiate with respect to $a$.

You are correct in using the chain rule. Remember when using the chain rule, you multiply the exponent of what is in the parentheses ($2$ in this case) which is where the $2$ comes from. Remember then that you must reduce the power of the parentheses by one. This yields $\sum_{i=1}^n 2*(y_i - {ae^{x_i^2}} -bx_i^3)^1$

The final step is to take the derivative of the inner function with respect to $a$. Since the inner function is $y_i - {ae^{x_i^2}} -bx_i^3$, the partial with respect to $a$ is $-e^{x_i^2}$ This is negative since the term is $- {ae^{x_i^2}}$ in the inner function.

Combining all that you have done so far you end up with:

$\sum_{i=1}^n 2*(y_i - {ae^{x_i^2}} -bx_i^3)^1*(-e^{x_i^2})$

The constants ($2$ and $-1$) can be removed to the front which gives your answer of $-2\sum_{i=1}^n (y_i - {ae^{x_i^2}} -bx_i^3)e^{x_i^2}$