Find dy/dx of : Y = ln ( x * (2x + 5)^1/3 ) / 7x^2 + 1
So to put that in words you have the function y = natural log of x times the cubed root of 2x+5 divided by 7 x squared + 1.
So far I have it broken down as follows:
I re wrote it up as ln (x) + ln (2x+5)^1/3 - ln (7x^2 + 1) using properties of logarithms.
I am having problems now with the - ln (7x^2 +1) part.
Before I get to far ahead of myself am I even headed in the right direction for what the problem is asking?
Basically I take finding dy/dx meaning say taking the function y = 3x^2.
Its derivative would be 6x (dx) and dy/dx would just be 6x.
Thats my first problem I'm working on.
I have two more that I have solved and am looking to make sure they are correct.
1) the indefinite integral of (x^4 / 3 - x^5) dx
For this one I set U = 3 - X^5 so that makes du = -4x^4.
So to balance I need to include a -4 in the numerator and subsequently add a factor of -1/4 outside of the integral. which makes it look like
-1/4 indef integral of (x^4) / (3 - x^5)
So for this one my final answer looks like -1/4 ln (3 - x^5) + C
2) the indefinite integral of sec(3x) dx
u = 3x
du = 3(dx)
du/3 = dx
= 1/3 * indef integral csc(u) du
= 1/3 * ln [ sec(u) + tan(u) ] + C
= 1/3 * ln [ sec(3x) + tan(3x) ] + C
So thats where I'm at, any help would be appreciated !!