Question

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

And please explain how you got your answer. I have no knowledge on how to get the answer AT ALL.

Answer 1

We're given the following equation:

`y=(x^2+2)^3(x^3+3)^2`

In order to find
`(dy)/(dx)` we must differentiate both sides of the equation.

Lets start differentiating the left side (
`y`):

`(d)/(dy) y=(1)dy=dy`

The
`(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:

`(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`):

`(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:

`(d)/(ds)[f(s)g(s)]= (d)/(ds)[f(s)]g(s)+(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

`(d)/(ds)[h(s)]= (d)/(ds)[g(f(s))] (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
`(dy)/(dx)`, and the answer is:

`(dy)/(dx) =6x(x^2+2)^2(x^3+3)^2+6x^2(x^3+3)(x^2+2)^3`