First take common factor of 3 of the terms inside the square brackets outsideOriginally Posted by becky
the square brackets
Now we also have a common factor of inside the square
brackets which we also take outside these brackets:
I am having trouble understanding a portion of the textbook example. It's seems to be mostly with the simplifing of the answer.
Find dy/dx if y = u^3 - 3u^2 + 1 and u = x^2 + 2
dy/du = 3u^2 - 6u and du/dx = 2x
it follows that dx/dy = (3u^2 - 6u)(2x). *I'm fine so far, keep going*
We are substituting x^2 + 2 for u in the expression for dy/dx giving
[3(x^2+2)^2 - 6(x^2+2)](2x)
*I understand everything above but the next step confuses me*
I don't know what steps they took to arrive at that equation. Could you possible show me the 'longer' process?
It goes on to give:
Hello, Becky (and anyone listening in) . . .
You might as well get used to simplifying messy derivatives.
. . It comes up quite often.
Example: Find the derivative of 2x-3)^2(x+5)^4" alt="f(x)\:=\2x-3)^2(x+5)^4" />
Product Rule: . 2x-3)^2\cdot4(x+5)^3 + 2(2x-3)\cdot2\cdot(x+5)^4" alt="f'(x)\:=\2x-3)^2\cdot4(x+5)^3 + 2(2x-3)\cdot2\cdot(x+5)^4" />
We have: . ... You'd probably stop here, right? *
Look at the common factors: .
Factor them out: .
. . and we have: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Your next question is (naturally) "Do we haaafta?"
(I'm not laughing at you . . . Years ago, I asked the same question.)
Normally, we take derivatives for a purpose, not just for exercise.
A common problem is to find the extreme values of a function.
We would set the derivative equal to zero and solve for
With your original form: .
. . you might multiply it out and stare at the fifth-degree equation for days.
But recall the procedure for solving an equation of any degree:
. . Factor the expression, set each factor equal to zero, and solve.
With the factored form, we have: .
. . and we can "eyeball" the solutions: .