Question

What is dy/dx if y = (x^2 + 2)^3 (x^3 + 3)^2

Answer 1

We're given the following equation:
$y=(x^2+2)^3(x^3+3)^2$

In order to find $\frac{dy}{dx}$ we must differentiate both sides of the equation.

Lets start differentiating the left side ($y$):
$\frac{d}{dy} y=(1)dy=dy$

The $\frac{d}{dy}$ simply servers to let us know we're differentiating whatever follows it (in this case $y$) with respect to $y$.

What we used to get the result is called the "power rule for differentiation" it states the following:
$\frac{d}{ds} s^{n}=ns^{n-1}$

In which $s$ is any variable (in the previous case $y$) and $n$ is any constant (in the previous case this $n=1$).

Now we'll differentiate the right side of the equation ($(x^2+2)^3(x^3+3)^2$):
$\frac{d}{dx}(x^2+2)^3(x^3+3)^2=[6x(x^2+2)^2(x^3+3)^2+6x^2(x^3+3)(x^2+2)^3]dx$

What we did to differentiate the right side was, first, apply something called "product rule" for differentiation, it states the following:
 $\frac{d}{ds}[f(s)g(s)]= \frac{d}{ds}[f(s)]g(s)+\frac{d}{ds}[g(s)]f(s)$

In which $f(s)$ and $g(s)$ are arbitrary functions of an arbitrary variable ($s$) (in this case $f(s)=f(x)=(x^2+2)^3$ and $g(s)=g(x)=(x^3+3)^2$).

After that we applied something called "chain rule" for differentiation, which states the following:
if $h(s)=g(f(s))$, then
$\frac{d}{ds}[h(s)]= \frac{d}{ds}[g(f(s))] \frac{d}{ds}[f(s)]$

Finally, the $dx$ we introduced as a factor after differentiating the right side (we also did it with the left side but with a $dy$) is a consequence of the chain rule, it is always done.

Finally, equaling both differentiated sides of the equation we have:
$dy=[6x(x^2+2)^2(x^3+3)^2+6x^2(x^3+3)(x^2+2)^3]dx$

We solve for $\frac{dy}{dx}$, and the answer is:
$\frac{dy}{dx} =6x(x^2+2)^2(x^3+3)^2+6x^2(x^3+3)(x^2+2)^3$