I dont understand where my mistake is. Can someone please help??! It involves derivatives.
I had to find the derivative of the original problem.
Did you try quotient rule?
Second step is wrong.
I used the product rule for the numerator and then I used the quotient rule for the simplified fraction.
What was wrong?
You can't just apply product rule for numeration while there is something not constant in denominator, you have to apply quotient right there.
multiply out the top, then use quotient rule
^
Oh okay. I'm gonna try that. So for future problems I can't combine different derivative rules in one problem?
well, roughly speaking, you shouldn't apply rule only to certain part of function, if you need to apply rule, you have to apply whole function.
I'm learning derivatives too, so I'm not an expert, but what I usually do is algebraically manipulate the function until I can use either a. a rule or b. a common derivative
softball if you want i can go over this problem with you on virtual whiteboard it will be more convenient
thanks guys for helping! I tried redoing the problem, but I got stuck on one of the last steps.
do you still need help or are you done ?
the textbook says that the answer is 3/x^4 I can see that I came close to getting that answer, but I am stumped on what to do with the expression that is in the parenthesis on the numerator.
you're taking derivative wrong if i am not wrong. i don't understand what you did in line 4 of your picture
Third line, just simplify right there: \[\dfrac{x^3-1}{x^3}=\dfrac{x^3}{x^3}-\dfrac{1}{x^3} = 1-\dfrac{1}{x^3}\]
So then does that mean that \[1-\frac{ 1 }{ x^{3}}\] is the answer? Or do I have to continue using the quotient rule to finish it?
In first three lines, you just simplify the function, so taking derivative would be easier. (y' in second and third line is supposed to be just y.) So now you take derivative of \(1-\dfrac{1}{x^3}\)
OHHHHH I just got it!!!!!
OMG! The lightbulb just went off! LOL :)
Thanks so much for your help!! You are awesome! :)
Glad we helped!
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