Express y=[4x-x^3]/[3x^2+2)^2 as a product and then differentiate. Simplify your answer using positive exponents. Help Please! ty
y = (4x - x^3)(3x^2 + 2)^-2
dy/dx = (4x - x^3)(-2)(3x^2+2)^-3(6x) + (3x^2+2)^-2 (4-3x^2)
= (3x^2+2)^-3 [ -12x(4-x^3) + (3x^2+2)(4-3x^2)]
= (3x^2 + 2)^-3 (3x^4 + 6x^2 - 48x + 8)
= (3x^4 + 6x^2 - 48x + 8)/(3x^2 + 2)^3
To express the equation y = [4x - x^3] / [(3x^2 + 2)^2] as a product, you can rewrite it as:
y = (4x - x^3) / (3x^2 + 2)^2
Now, let's find the common factors in the numerator and denominator to simplify the expression further.
Step 1: Factor out x in the numerator:
y = x(4 - x^2) / (3x^2 + 2)^2
Step 2: Recognize that the numerator is the difference of squares:
y = x(2 - x)(2 + x) / (3x^2 + 2)^2
Step 3: Now, the equation can be expressed as a product:
y = x(2 - x)(2 + x) / [(3x^2 + 2)(3x^2 + 2)]
To differentiate the equation, we will use the quotient rule. The quotient rule states that if you have a function u(x) divided by another function v(x), with u'(x) and v'(x) as their respective derivatives, then the derivative of the quotient is given by:
[d/dx (u(x) / v(x))] = [v(x)(u'(x)) - u(x)(v'(x))] / (v(x))^2
In this case, u(x) = x(2 - x)(2 + x) and v(x) = (3x^2 + 2)(3x^2 + 2).
Let's differentiate the equation step by step:
Step 1: Determine u'(x):
u'(x) = [d/dx (x(2 - x)(2 + x))] = (2 - x)(2 + x) + x(2 + x)(-1) + x(2 - x)(1) = (2 - x)(4 + 2x - x) - x(2 + x) = (2-x)(3+x)
Step 2: Determine v'(x):
v'(x) = [d/dx ((3x^2 + 2)(3x^2 + 2))] = [(3x^2 + 2)(6x) + (6x)(3x^2 + 2)] = 12x(3x^2 + 2)
Step 3: Plug in the values into the quotient rule equation:
[d/dx (y)] = [v(x)(u'(x)) - u(x)(v'(x))] / (v(x))^2
= [(3x^2 + 2)(3x^2 + 2)(2-x)(3+x) - x(2 - x)(2 + x)(12x)] / [(3x^2 + 2)(3x^2 + 2)]^2
Simplifying the expression would involve expanding and collecting like terms, but as you specified to simplify using positive exponents, we'll leave it in this form:
[d/dx (y)] = [(3x^2 + 2)(3x^2 + 2)(2-x)(3+x) - x(2 - x)(2 + x)(12x)] / [(3x^2 + 2)(3x^2 + 2)]^2
This is the derivative of y with respect to x.
To express the given expression as a product, we can factor the numerator and denominator separately.
Let's start by factoring the numerator, which is 4x - x^3. We can see that there is a common factor of x:
y = [x(4 - x^2)] / [3x^2 + 2)^2
Next, let's factor the denominator, which is (3x^2 + 2)^2. This is already in factored form, so we leave it as is.
Now, we can express the given expression as a product:
y = [x(4 - x^2)] / [(3x^2 + 2)^2]
= x(4 - x^2) / (3x^2 + 2)^2
Now, let's differentiate the expression. We can use the quotient rule for differentiation, which states that for functions u(x) and v(x):
(d/dx)(u(x) / v(x)) = (v(x) * u'(x) - u(x) * v'(x)) / v(x)^2
Applying the quotient rule to our expression:
y' = [(3x^2 + 2)^2 * (4 - x^2)' - (4 - x^2) * (3x^2 + 2)^2'] / (3x^2 + 2)^4
Now, let's find the derivatives of the individual terms:
(4 - x^2)' = -2x
(3x^2 + 2)^2' = 2(3x^2 + 2)(6x)
Substituting these derivatives back into the differentiation expression:
y' = [(3x^2 + 2)^2 * (-2x) - (4 - x^2) * 2(3x^2 + 2)(6x)] / (3x^2 + 2)^4
Now, let's simplify this expression using positive exponents:
y' = [-2x(3x^2 + 2)^2 - 2(4 - x^2)(3x^2 + 2)(6x)] / (3x^2 + 2)^4
So, the derivative of y with respect to x in its simplified form is:
y' = [-2x(3x^2 + 2)^2 - 2(4 - x^2)(3x^2 + 2)(6x)] / (3x^2 + 2)^4