Find f'(x): (abs(((x^2)*((3x+2) ^(1/3)))/((2x-3)^3))
So I thought I should split it into two parts: A and B.
A) (x^2)/((2x-3)^2)
B) ((3x+2)^(1/3))/(2x-3)
My plan was to differentiate each part and then multiply them together, since multiplying their denominators together gives me the original denominator power (4).
I end up getting for A) (-12x^2 + 18x)/((2x-3)^4)
For B, I have something which seems much more complex, because of the negative fractional exponents, and this is where I'm stuck. I want to simplify, but I'm not sure how to without screwing everything up. This is my last step for B:
((3x+2)^(-2/3)*(2x-3) - 4(x-3)*((3x+2)^(1/3)))/((2x-3)^2)
I really appreciate the help. And also, as for the absolute value part of the question, well I haven't even thought about that- I'm stuck on the easy part itself. -_-
Thank you!
X^2(3x-2)^-2/3/(2x-3)^3 - 2x(x+3)(3x+2)^1/3/(2x-3)^4
The strategy is interesting, since you'll have to apply the quotient rule to each part, then the product rule.
Your value in A is correct, but note that
(-12x^2 + 18x)/((2x-3)^4)
= -6x(2x-3)/(2x-3)^4
= -6x/(2x-3)^3
B' = -(4x+7)/((2x-3)^2 (3x+2)^(2/3))
Now you have to evaluate
A'B + AB'
What a mess.
I'd rather just work with the original expression. Unfortunately, the parentheses are not balanced. Try
f(x) = |uv/w^3| if that is what is meant. Let us know what u,v,w are. I think it's clear, but I'm not sure where the || applies.
To find the derivative of the given function f(x), you have correctly split it into two parts, A and B. Now, let's go through each part and find their derivatives.
Part A:
A = (x^2) / ((2x-3)^2)
To differentiate part A, you can use the quotient rule:
if f(x) = u(x)/v(x), then f'(x) = (u'(x)v(x) - u(x)v'(x)) / [v(x)]^2
In this case, u(x) = x^2 and v(x) = (2x-3)^2.
Differentiating u(x) with respect to x gives u'(x) = 2x.
Differentiating v(x) with respect to x gives v'(x) = 2(2x-3).
Applying the quotient rule, we have:
A' = [(u'(x)v(x) - u(x)v'(x))] / [v(x)]^2
= [(2x)(2x-3)^2 - (x^2)(2(2x-3))] / [(2x-3)^2]^2
= [-12x^2 + 18x] / [(2x-3)^4]
So, you have correctly found the derivative of part A as (-12x^2 + 18x) / (2x-3)^4.
Now, let's move on to part B.
Part B:
B = ((3x+2)^(1/3)) / (2x-3)
To differentiate part B, we can use the quotient rule again.
But before we differentiate, let's simplify the expression inside the numerator by rationalizing the denominator.
Multiply the numerator and denominator of the expression ((3x+2)^(1/3)) by the cube root of ((3x+2)^2):
B = (((3x+2)^(1/3))*(3x+2)^2) / ((2x-3)*((3x+2)^(2/3)))
Now, the expression inside the numerator becomes ((3x+2)^(1/3)) * ((3x+2)^(2/3)), which is equal to (3x+2)^(1/3 + 2/3), simplifying to (3x+2).
Thus, we have:
B = (3x+2) / ((2x-3)*((3x+2)^(2/3)))
Now, let's differentiate B using the quotient rule.
In this case, u(x) = 3x+2 and v(x) = (2x-3)*((3x+2)^(2/3)).
Differentiating u(x) with respect to x gives u'(x) = 3.
Differentiating v(x) with respect to x requires the product rule:
v'(x) = [2((3x+2)^(2/3))] + [(2x-3)* (2/3)*((3x+2)^(2/3-1))* (3)].
Let's simplify v'(x):
v'(x) = [2((3x+2)^(2/3))] + [(2x-3)* (2/3)*((3x+2)^(-1/3))* (3)]
= [2(3x+2)^(2/3)] + [(2x-3)* (2/3)* (3) / ((3x+2)^(1/3))].
Now, applying the quotient rule:
B' = [(u'(x)v(x) - u(x)v'(x))] / [v(x)]^2
= [(3 * (2x-3)*((3x+2)^(2/3))) - ((3x+2) * (2(3x+2)^(2/3)) + (2x-3)* (2/3)* (3) / ((3x+2)^(1/3))) / [((2x-3)*((3x+2)^(2/3)))^2]
= [6x - 9 - ((6x+4)*(2/3)*(3)) / ((3x+2)^(1/3))] / [((2x-3)*((3x+2)^(2/3)))^2]
Note: The absolute value part of the equation (abs()) can be addressed by considering two scenarios, one where the expression inside the absolute value is positive and the other where it is negative. However, since you only requested assistance with the differentiation part, I have focused on that. If you need help with the absolute value part as well, please let me know.
In conclusion, you have successfully found the derivative of both parts A and B. For part A, the derivative is (-12x^2 + 18x) / (2x-3)^4, and for part B, the derivative is [6x - 9 - ((6x+4)*(2/3)*(3))] / [((2x-3)*((3x+2)^(2/3)))^2].