Our final derivatives are thus
$$ \frac{\partial f}{\partial c} = \frac{\partial f}{\partial d} \frac{\partial d}{\partial c} = \frac{1}{2\sqrt{c}}, $$ $$ \frac{\partial f}{\partial b} = \frac{\partial f}{\partial c} \frac{\partial c}{\partial b} = \frac{1}{2\sqrt{c}}, $$ $$ \frac{\partial f}{\partial a} = \frac{\partial f}{\partial c} \frac{\partial c}{\partial a}+ \frac{\partial f}{\partial b} \frac{\partial b}{\partial a} = \frac{1+\exp{a}}{2\sqrt{c}}, $$and finally
$$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial a} \frac{\partial a}{\partial x} = \frac{x(1+\exp{a})}{\sqrt{c}}, $$which is just
$$ \frac{\partial f}{\partial x} = \frac{x(1+b)}{d}, $$and requires only three operations if we can reuse all intermediate variables.