For these problems find y.
1. y'= 3x-5
so basically I'm suppose to go backwards this time instead of finding the derivative I'm finding the original y?
correct. Use the power rule
Since d/dx x^n = n x^(n-1)
That is, when taking the derivative, you divide by the power and subtract one.
To go backwards, reverse the operations -- add one to the power and then divide by it.
If y' = x^n, then y = 1/(n+1) x^(n+1)
y = 3/2 x^2 - 5x + c
where c is an arbitrary constant (whose derivative is just 0)
Ok! I was able to do a few problems like that^
What about y'= 1/x^2+1/x ?
come on, man. Don't forget your Algabra I now that you're taking calculus!
y' = 1/x^2 + 1/x = x^-1 + x^-1
The x^-2 is easy -- y = 1/-1 x^-1 = -1/x
The y' = 1/x is a bit different. It should look familiar, though. Recall that
d/dx lnx = 1/x
so, if y' = 1/x, y = lnx
So this one winds up as -1/x + lnx + c
Yes, in this problem, you are given the derivative of y (denoted as y'), and you need to find the original function y.
To find y, you can use the process of integration. Integration is the reverse operation of differentiation. By integrating y' with respect to x, you can determine y.
To solve the problem, follow these steps:
1. Start with the given derivative: y' = 3x - 5.
2. Integrate both sides of the equation with respect to x. Integrating y' gives you the original function y: ∫y' dx = ∫(3x - 5) dx.
3. Apply the power rule of integration to ∫(3x - 5) dx. This rule states that the integral of x^n with respect to x is (1/n+1) * x^(n+1), where n is any real number except -1. Integrating each term separately:
∫3x dx = (3/2) * x^2 + C1, where C1 is the constant of integration for the first term.
∫(-5) dx = -5x + C2, where C2 is the constant of integration for the second term.
4. Combine the integrated terms with their respective constants of integration to find the original function y:
y = (3/2) * x^2 - 5x + C, where C = C1 + C2 is the overall constant of integration.
Therefore, the solution for y, given y' = 3x - 5, is y = (3/2) * x^2 - 5x + C, where C is a constant.