So, I have a bunch of derivative problems for homework and the only resource I have is the textbook, since we don't have notes. I have just started and want to make sure I'm doing them correctly. Our teacher said we do not have to simplify them either, so I'm also confused if I'm stopping at the right point.
Differentiate.
1. Y(u)= (u^-2 + u^-3)(u^5 - 2u^2)
So, f(x)= (u^-2 + u^-3)
g(x)= (u^5 - 2u^2)
power rule:
= (-2u^-3 + -3u^-4)(5u^4 - 4u)
product rule:
= (-2u^-3 + -3u^-4) * D(5u^4 - 4u) + (5u^4 - 4u) * D(-2u^-3 + -3u^-4)
D(5u^4 - 4u)= 20u^3
D(-2u^-3 + -3u^-4)= 6u^-4 + 12u^-5
2. y = (x + 1)/(x^3 + x - 2)
y-prime=
[(x^3 + x - 2) * D(x + 1) - (x + 1) * D(x^3 + x - 2)]/(x^3 + x - 2)^2
D(x + 1)= x ?
D(x^3 + x - 2)= 4x^2 ?
Please help. I need to understand these, as I have plenty to do!
1. My first-line derivative would be
Y' (u) = (u^-2 + u^-3)(5u^4 - 4u) + (u^5-2u^2)(-2u^-3 - 3u^-4)
If I were testing my students to see if they master the product rule, this is the line I would be looking for.
The rest is "window dressing", that is, simplifying it.
If you want a very simple explanation of the product rule, it would be ...
if y = (u)(v)
then y' = uv' + vu'
2. my first-line derivative of this quotient rule is
y' = [ (x^3+ x - 2)(1) - (x+1)(3x^2 + 1) ]/(x^3+x-2)^2
in short,
if y = u/v
then y' = (vu' - uv')/v^2
Great job so far! Let's go through your solutions step by step and see if we can make any corrections.
1. Y(u) = (u^-2 + u^-3)(u^5 - 2u^2)
You correctly identified the two functions: f(u) = u^-2 + u^-3 and g(u) = u^5 - 2u^2.
Applying the power rule to f(u), we have:
f'(u) = -2u^-3 + -3u^-4
And applying the power rule to g(u), we have:
g'(u) = 5u^4 - 4u
Now, using the product rule, we can find the derivative of Y(u):
Y'(u) = f'(u) * g(u) + f(u) * g'(u)
Substituting the expressions we derived earlier:
Y'(u) = (-2u^-3 + -3u^-4) * (u^5 - 2u^2) + (u^-2 + u^-3) * (5u^4 - 4u)
Simplifying this expression gives:
Y'(u) = (-2u^-2 + -3u^-3) * (u^5 - 2u^2) + (u^-1 + u^-2) * (5u^4 - 4u)
Now, let's move on to the second problem.
2. y = (x + 1)/(x^3 + x - 2)
To find the derivative of y with respect to x, we can apply the quotient rule:
The quotient rule states that for a function u(x)/v(x), the derivative is given by:
(u'(x)v(x) - u(x)v'(x))/v(x)^2
Let's apply the quotient rule to our function y:
y' = [(x^3 + x - 2) * (D(x + 1)) - (x + 1) * (D(x^3 + x - 2))]/(x^3 + x - 2)^2
D(x + 1) is correct. The derivative of x + 1 with respect to x is simply 1.
Now, let's find the derivative of x^3 + x - 2. Using the power rule, we have:
D(x^3 + x - 2) = 3x^2 + 1
Therefore, we have:
y' = [(x^3 + x - 2) * 1 - (x + 1) * (3x^2 + 1)]/(x^3 + x - 2)^2
Simplifying the expression gives us the derivative of y with respect to x.
I hope this helps! Remember to always double-check your work and don't hesitate to ask if you have any further questions. Good luck with your homework!