How to calculate the partial derivative of function within another function?

Question

I need someone help to calculate the partial derivative. Following is the question

$$R=\frac{ba^2}{(b+k)^2}-\textrm{Cost}\left(\frac{ba}{b+k}\right)$$

Derivative of the first term has been calculated as following $$\frac{\partial}{\partial b}\frac{ba^2}{(b+k)^2}=\frac{(b+k)^2a^2-2a^2b(b+k)}{(b+k)^4}=\frac{a^2k-ba^2}{(b+k)^3}$$

Now, what will be the partial derivative of the following (that is the second term of $R$)? $$\frac{\partial}{\partial b}\textrm{Cost}\left(\frac{ba}{b+k}\right)=?$$

$$\textrm{Cost}(x)={\lambda_2x^2}+{\lambda_1x}-{\lambda_0}$$

Note: here, I need to calculate the partial derivative of the following two cases e.g., different values of $x$ variable

1. $$x=a$$

2. $$x=\frac{ba}{b+k}$$

$$\frac{\partial R}{\partial b}=\frac{\partial}{\partial b}\left[\frac{ba^2}{(b+k)^2}-\textrm{Cost}\left(\frac{ba}{b+k}\right)\right]=?$$ Please help to me calculate the deravitive in both cases, I will need both solutions.

Answer 1

Use the chain rule

$$ \frac{\partial C}{\partial b} = \frac{dC}{dx}\frac{\partial x}{\partial b} $$

where $C$ is the cost function.

You then have

$$ \frac{dC}{dx} = 2\lambda_2x + \lambda_1 $$

and

$$ \frac{\partial x}{\partial b} = \frac{\partial}{\partial b}\left(\frac{ab}{b+k}\right) = \frac{ak}{(b+k)^2} $$

Now multiply them and plug in $x$ in terms of $b$