Why does my textbook solve an equation with the following term:
Sin x^2 y^2
As d/dx (Sin x^2 y^) x d/dx (x^2 y^2)
Are you "solving an equation", as you state, or is the example actually showing, perhaps, how to apply the Chain Rule to the process of finding a derivative? If the former, then we'll need
loads more information, such as the original equation, and all of the steps that led up to the one you post. If the latter, then:
The Chain Rule, effectively, says to "differentiate from the outside, inwards".
Remember when you evaluated composed functions, like "find the value of f(g(h(x))) when x = 6"? You had to start from the inside, and work outwards; you'd plug "6" in for "x" in h(x), then plug that in for "x" in g(x), and then plug
that in for "x" in f(x).
For the Chain Rule, it works the other way 'round. You differentiate the outer layer, times (you differentiate the next layer), times (you differentiate the
next layer), and so forth, until you finally drill down to the "x".
So, for instance, if you were given:
. . . . .\(\displaystyle \mbox{Differentiate }\, y\, =\, \sqrt{\strut \ln\big(\cos(x^2)\big)\,}\)
...you wouldn't start inside with the "x"; you'd work from the outside inward. What's the first function you get to, starting on the outside? You first come to the square root; so you'd differentiate the square root (being a one-half power), getting a fractional something-or-other, but with the insides of the square root being unchanged. Then you'd multiply by... What's the next function? The first thing you get to, inside the radical, is not the square of the x or the cosine; the x^2 is inside the cosine, and the cosine is inside the log. So the log is the next then you get to. So put a multiplication "dot" after the square-root derivative stuff, and differentiate the log; you'll get a fractional something-or-other, with the insides of the (something-or-other) being the unchanged argument of the log (that is, the cosine of the square). Then you do another multiplication "dot", and differentiate the cosine. And so forth:
. . . . .\(\displaystyle y\, =\, \sqrt{\strut \ln\big(\cos(x^2)\big)\,}\)
. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{d}{dx}\, \sqrt{\strut \ln\big(\cos(x^2)\big)\,}\)
. . . . . . . .\(\displaystyle =\, \dfrac{1}{2}\, \dfrac{1}{\sqrt{\strut \ln\big(\cos(x^2)\big)\,}}\, \cdot\, \dfrac{d}{dx}\,\ln\big(\cos(x^2)\big)\)
. . . . . . . .\(\displaystyle =\, \dfrac{1}{2}\, \dfrac{1}{\sqrt{\strut \ln\big(\cos(x^2)\big)\,}}\, \cdot\, \dfrac{1}{\cos(x^2)}\, \cdot\, \dfrac{d}{dx}\, \cos(x^2)\)
. . . . . . . .\(\displaystyle =\, \dfrac{1}{2}\, \dfrac{1}{\sqrt{\strut \ln\big(\cos(x^2)\big)\,}}\, \cdot\, \dfrac{1}{\cos(x^2)}\, \cdot\, -\sin(x^2)\, \cdot\, \dfrac{d}{dx}\, x^2\)
. . . . . . . .\(\displaystyle =\, \dfrac{1}{2}\, \dfrac{1}{\sqrt{\strut \ln\big(\cos(x^2)\big)\,}}\, \cdot\, \dfrac{1}{\cos(x^2)}\, \cdot\, -\sin(x^2)\, \cdot\, 2x\)
Once you're down to the "x", you're done.
Now apply the same process to the expression (not "equation") you were given.
