derivative using chain rule $h(x)c(y)exp(w(y)t(x))$
the derivative of this with respect to y is $h(x)c'(y)exp(w(y)t(x))+h(x)c(y)w'(y)t(x)exp(w(y)t(x))$
I am having trouble with the derivative of this term because using the chain rule with three functions is throwing me off.
so $\frac{d}{dy} [h(x)c'(y)exp(w(y)t(x))+h(x)c(y)w'(y)t(x)exp(w(y)t(x))]$
 A: Let $f(x,y)= h(x)c(y)$ & $g(x,y)=\omega (y)t(x)$. Then your expression is:
$$= f(x,y)e^{g(x,y)}$$
It is important to note that:
$$\frac{\partial}{\partial y} f(x,y)= f'(x,y)= h(x) c'(y)$$
$$\frac{\partial}{\partial y} g(x,y)= g'(x,y)= t(x) \omega '(y)$$
Taking the first derivative:
$$\frac{\partial }{\partial y}f(x,y)e^{g(x,y)}=f'(x,y)e^{g(x,y)}+f(x,y)g'(x,y)e^{g(x,y)}$$
The second derivative would be:
$$\frac{\partial ^2}{\partial y^2}(f(x,y)e^{g(x,y)}=\frac{\partial}{\partial y}f'(x,y)e^{g(x,y)}+f(x,y)g'(x,y)e^{g(x,y)})$$
$$=\frac{\partial }{\partial y}(f'(x,y)e^{g(x,y)})+\frac{\partial}{\partial y}f(x,y)g'(x,y)e^{g(x,y)}$$
The first term would be:
$$\frac{\partial }{\partial y}(f'(x,y)e^{g(x,y)})= f''(x,y)e^{g(x,y)}+f'(x,y)g'(x,y)e^{g(x,y)}$$
The second term is:
$$\frac{\partial}{\partial y}f(x,y)g'(x,y)e^{g(x,y)}=f'(x,y)g'(x,y)e^{g(x,y)}+f(x,y)g''(x,y)e^{g(x,y)}+f(x,y)(g'(x,y))^2e^{g(x,y)}$$
Altogether the second derivative is then:
$$\frac{\partial ^2}{\partial y^2}(f(x,y)e^{g(x,y)}=f''(x,y)e^{g(x,y)}+f'(x,y)g'(x,y)e^{g(x,y)}+f'(x,y)g'(x,y)e^{g(x,y)}+f(x,y)g''(x,y)e^{g(x,y)}+f(x,y)(g'(x,y))^2e^{g(x,y)}$$
Or in terms of the functions you started with:
$$=h(x)c''(y)e^{\omega (y) t(x)}+2h(x)c'(y)\omega '(y) t(x)e^{\omega (y) t(x)}+h(x)c(y)\omega ''(y) t(x)e^{\omega (y) t(x)}+h(x)c''(y)(\omega '(y) t(x))^2e^{\omega (y) t(x)}$$
