We're given the following equation:
[tex]y=(x^2+2)^3(x^3+3)^2[/tex]
In order to find [tex] \frac{dy}{dx} [/tex] we must differentiate both sides of the equation.
Lets start differentiating the left side ([tex]y[/tex]):
[tex] \frac{d}{dy} y=(1)dy=dy[/tex]
The [tex] \frac{d}{dy} [/tex] simply servers to let us know we're differentiating whatever follows it (in this case [tex]y[/tex]) with respect to [tex]y[/tex].
What we used to get the result is called the "power rule for differentiation" it states the following:
[tex] \frac{d}{ds} s^{n}=ns^{n-1} [/tex]
In which [tex]s[/tex] is any variable (in the previous case [tex]y[/tex]) and [tex]n[/tex] is any constant (in the previous case this [tex]n=1[/tex]).
Now we'll differentiate the right side of the equation ([tex](x^2+2)^3(x^3+3)^2[/tex]):
[tex]\frac{d}{dx}(x^2+2)^3(x^3+3)^2=[6x(x^2+2)^2(x^3+3)^2+6x^2(x^3+3)(x^2+2)^3]dx[/tex]
What we did to differentiate the right side was, first, apply something called "product rule" for differentiation, it states the following:
[tex] \frac{d}{ds}[f(s)g(s)]= \frac{d}{ds}[f(s)]g(s)+\frac{d}{ds}[g(s)]f(s) [/tex]
In which [tex]f(s)[/tex] and [tex]g(s)[/tex] are arbitrary functions of an arbitrary variable ([tex]s[/tex]) (in this case [tex]f(s)=f(x)=(x^2+2)^3[/tex] and [tex]g(s)=g(x)=(x^3+3)^2[/tex]).
After that we applied something called "chain rule" for differentiation, which states the following:
if [tex]h(s)=g(f(s))[/tex], then
[tex] \frac{d}{ds}[h(s)]= \frac{d}{ds}[g(f(s))] \frac{d}{ds}[f(s)] [/tex]
Finally, the [tex]dx[/tex] we introduced as a factor after differentiating the right side (we also did it with the left side but with a [tex]dy[/tex]) is a consequence of the chain rule, it is always done.
Finally, equaling both differentiated sides of the equation we have:
[tex]dy=[6x(x^2+2)^2(x^3+3)^2+6x^2(x^3+3)(x^2+2)^3]dx[/tex]
We solve for [tex] \frac{dy}{dx} [/tex], and the answer is:
[tex] \frac{dy}{dx} =6x(x^2+2)^2(x^3+3)^2+6x^2(x^3+3)(x^2+2)^3[/tex]
