User:Mikasen

Triple product rule of differentiation:

Let $$x^3y^2+y=\sin x$$

$${d \over dx}(x^3y^2)+{dy \over dx}={d \over dx}(\sin x)$$

$$3x^2y^2+2x^3y{dy \over dx}+{dy \over dx}=\cos x$$

Let $$y=uvw$$, where $$u=f(x)$$, $$v=g(x)$$, $$w=h(x)$$

We have $$y+\delta y=(u+\delta u)(v+\delta v)(w+\delta w)$$

$$y+\delta y=uvw+u\delta v\delta w+v\delta u\delta w+w\delta u\delta v+uv\delta w+uw\delta v+vw\delta u+\delta u\delta v\delta w$$

$${\delta y \over \delta x}={u\delta v\delta w \over \delta x}+{v\delta u\delta w \over \delta x}+{w\delta u\delta v \over \delta x}+{uv\delta w \over \delta x}+{uw\delta v \over \delta x}+{vw\delta u \over \delta x}+{\delta u\delta v\delta w \over\delta x}$$

As $$\delta x\rightarrow 0$$, the terms $${u \delta v \delta w \over \delta x}$$, $${v\delta u\delta w \over \delta x}$$, $$w\delta u\delta v \over \delta x$$ and $$\delta u\delta v\delta w \over \delta x$$ become negligible.

Hence, $${dy \over dx}=uv{dw \over dx}+uw{dv \over dx}+vw{du \over dx}$$ (shown)