Re: Distance Learning and Math Courses
@32, here's my best attempt at explaining the concepts you mentioned:
As far as the problem you presented, you would use the quotient rule to solve. The rule states:
The derivative of a rational function is calculated by FPrime*G-F*GPrime all divided by G^2,
where F(X) = the numerator and G(X) = the denominator and you are dividing by 0.
Knowing this, you can solve for both FPrime and GPrime by deriving the numerator and denominator. At that point, it's just a matter of plugging things and doing the algebra. Just to make sure we both are getting the same answers, I contend that:
FPrime = d/dx(12x^2-19x) = 24x-19
GPrime = d/dx(14x) = 14
You now have all of what you need to plug in and solve using the quotient rule. It's just algebraic work.
Regarding the chain rule, it's actually far more simple than the definition you provided is leading you to believe. All it states is:
The derivative of F(G(X)), or, in plain terms, a function composed of another function, is equal to the derivative of F(X), or the outside function, times the derivative of G(X), or in the inside function.
Let's say we want to derive (4x^3+2x)^2. Note that we are raising the quantity of 4x^3+2x to the second power, as delineated by the parentheses. Whatever is inside of the parentheses is its own function, but we are raising that whole thing to an exponent, so we split this up into two functions. The chain rule is most apparent through exponents, so this is probably the most common type of problem you'll see next to trig derivatives. Now, we represent what I just stated as:
F(X) = ( G(X) )^2, where G(X) is just another function.
G(X) = (4x^3+2x)
It really helps to force parentheses around things to help make it clear which terms are part of a single function. In this case, the stuff inside the parentheses surrounding G(X) will eventually be raised to a second power, meaning whatever value G(X) produces will be squared.
Now, and this is also extremely important, when you are deriving F(X), you do not consider anything inside of the parentheses. F(X) was defined as a placeholder that squares whatever is inside of it. In other words, don't even think about G(X) yet, we're just focused on ( ... )^2, where ... is just stuff inside of F(X).
To make this even more apparent, let's just say that ( ... ) = u. Then we see that we are deriving u^2.
Just like you would do for x^2, deriving u^2 gives us 2u. Plugging ( ... ) back in for u, we now have 2( ... ). Plugging G(X) back in for ..., we now see that the derivative of the outside function is 2(4x^3+2). That is precisely how you derive the outside function, or F(X).
To solve for the derivative of the inside function, it's actually simple. Recall that we let G(X) be equal to 4x^3+2x. Deriving that gives us 12x^2+2.
Finally, we multiply both of these things together. So we'd have:
2(4x^3+2x)
*
(12x^2+2)
to give us:
2*(4x^3+2x0*(12x^2+2)
If you really wanted to, you could distribute all of that out, but it's needless algebra at this point. No use in simplify.
I can understand the confusion over how you define F and G. Try to remember that if you can place parentheses around terms which then are being altered by exponents, everything outside of the parentheses is your F. If we had multiplied all of this by 2, it would've looked like:
2(4x^3+2x)^2
so we'd be deriving 2( ... )^2, or 2u^2.
You'll also probably find it easier to understand all of this by using the substitutions like I did. It's really deriving the outer function people don't quite understand, but if you're not getting it, try replacing things with names that look similar to variables so it becomes more apparent when you're differentiating.
Let me know if any of that made sense.
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