partial derivative chain rules Suppose that there is $f(a,b)$. Also suppose that $b = g(a, \text{and some other variables})$.
By chain rule, it seems that $$\frac{\partial f}{\partial a} = \frac{\partial f}{\partial b}\frac{\partial b}{\partial a}+\frac{\partial f}{\partial a}\frac{\partial a}{\partial a}$$.
Is this accurate? 
Edit: If this is not true, how is $\frac{\partial f}{\partial a}$ and $\frac{\partial f}{\partial b}$ related?
Edit 2: OK. if $f(b,c)$ and $b=g(a, z, y, x...)$ and $c = a$. Now chain rule:
$$\frac{\partial f}{\partial a} = \frac{\partial f}{\partial b}\frac{\partial b}{\partial a} + \frac{\partial f}{\partial c}\frac{\partial c}{\partial a} = \text{this equals to the above}.$$ So, what's wrong with this?
 A: this equation is useful:
$$\frac{df}{da}=\frac{df}{db}\frac{db}{da}$$
Your original statement is not incorrect in thought but it is very confusing how you having it written.  Try using the following.
$$\frac{df}{dt}=\frac{df}{da}\frac{da}{dt}+\frac{df}{db}\frac{db}{dt}$$
and set t = a.  It will make it easier.  If you just do the way to have it written:
$$\frac{df}{da}=\frac{df}{da}+\frac{df}{db}\frac{db}{da}$$
which is confusing.  However:
$$f(a,b)=a^2+ab^2$$
$$\frac{df}{dt}=(2a+b^2)\frac{da}{dt}+(2ab)\frac{db}{dt}$$ Then $$\frac{df}{da}=2a+b^2+2ab\frac{db}{da}$$ is alot clearer and 100% correct.
Try it out.
I believe Maisam Hedyelloo was confused by the way it is written and I cannot blame him as I would have looked at it the same way.
A: First of all are you talking about total derivative or partial derivative? The total derivative of a function $f(x_1,...,x_n)$ is given by
$$df(x_1,...,x_n)=\frac{\partial f}{\partial x_1}dx_1+...+\frac{\partial f}{\partial x_n}dx_n$$
By definition the components in total derivative are named as partial derivative, such as $\frac{\partial f}{\partial x_i}$ is the partial derivative with respect to $x_i$. Your statement about the chain rule is correct and can be stated as
$$\frac{\partial f\big(a,b(a)\big)}{\partial a}=\frac{\partial f}{\partial a}+\frac{\partial f}{\partial b}\frac{db}{da}$$
By definition of partial derivative with respect to variable "a" other variables are not important (imagine them as fixed variables) if they don't depend on "a". To show it
$$\frac{\partial f\big(a,b(a,z),x,y\big)}{\partial a}=\frac{\partial f}{\partial a}+\frac{\partial f}{\partial b}\frac{db}{da}$$
