Find the derivative dy/dx for the following. Simplify your answer.
y=(ln(x^2))/(x^3)
I got (2/x x^3-3x^2 ln(x^2))/((x^3)^2) for the derivative is that correct? Or no?
It looks good, but could probably be massaged a bit. How about
(2-6logx)/x^4
Steve I got the answer that you have originally but got it marked wrong on the test so that's why I'm unsure.
that's the problem with complicated stuff like this. There are always several ways to write it. Bad area for short-answer responses.
To find the derivative of the given function, which is y = (ln(x^2))/(x^3), we can use the quotient rule of derivatives. The quotient rule states that if you have a function of the form y = f(x)/g(x), then the derivative is given by:
dy/dx = (g(x)f'(x) - f(x)g'(x))/(g(x))^2
Let's apply the quotient rule to find the derivative of y = (ln(x^2))/(x^3):
First, we'll find the derivative of the numerator, f(x), which is ln(x^2). The derivative of ln(u) is (1/u) * u', where u' is the derivative of u with respect to x.
Taking the derivative of ln(x^2), we have: f'(x) = (1/(x^2)) * (2x) = (2x/(x^2)) = 2/x.
Next, we'll find the derivative of the denominator, g(x), which is x^3. The derivative of x^n (where n is a constant) is n * x^(n-1).
Taking the derivative of x^3, we have: g'(x) = 3 * x^(3-1) = 3 * x^2 = 3x^2.
Now, we'll substitute our values into the quotient rule formula:
dy/dx = (g(x)f'(x) - f(x)g'(x))/(g(x))^2
= (x^3 * (2/x) - ln(x^2) * 3x^2) / (x^3)^2
= (2x^2 - 3x^2 * ln(x^2))/x^6
Finally, we can simplify the expression by factoring out an x^2 from the numerator:
dy/dx = (2x^2(1 - 3ln(x^2)))/x^6
= (2(1 - 3ln(x^2)))/x^4
Hence, the simplified derivative of y = (ln(x^2))/(x^3) is dy/dx = (2(1 - 3ln(x^2)))/x^4.