Partial derivative of square formula

Question

I want to know the partial derivative w.r.t '$b$' of the following equation:

$$1/2(a-b)^2$$

I have tried solving it down but my overall answer computed due to this derivative is incorrect.

Please help.

Answer 1

Applying the classical formula for the derivative of a composition of two functions $(f \circ g)(x) = f'(g(x)) g'(x)$ you get

$$ \frac{\partial}{\partial b} \left( \frac{1}{2} (a-b)^2 \right) = (a-b) \times (-1) = b - a $$

Answer 2

I think you want to calculate $$ \frac{\partial}{\partial b} \frac{1}{2(a-b)^2} $$.

As Dodo say, you have to use the chain rule. Here $f(x)=\frac{1}{2x^2}$ and $g(x)=a-b$ , so $f \circ g (x) = \frac{1}{2(a-b)^2}$

The chain rule say then:

$$ \frac{\partial}{\partial b} \frac{1}{2(a-b)^2} = \frac{-1}{4(a-b)^3} (-1) = \frac{1}{4(a-b)^3} $$.