Need an answer to this Calculus Question. I keep getting this type of question wrong. Thanks.
Differentiate the Function:
y = (3x-1)^2(2-x^4)^2
y =
do is as a uv problem
d(uv)=vdu+udv
y'=2(3x-1)(3) [2-x^4)^2 + 2(2-x^4)(-4x^3)(3x-1)^2
have fun gathering terms.
To differentiate the given function, y = (3x-1)^2(2-x^4)^2, you can use the product rule and the chain rule. Here is the step-by-step explanation:
Step 1: Apply the product rule:
The product rule states that if y = u * v, where u and v are functions of x, then the derivative of y with respect to x is given by:
dy/dx = u * dv/dx + v * du/dx
In our case, u = (3x-1)^2 and v = (2-x^4)^2. So, applying the product rule, we get:
dy/dx = (3x-1)^2 * d(2-x^4)^2/dx + (2-x^4)^2 * d(3x-1)^2/dx
Step 2: Differentiate each term separately:
To differentiate (2-x^4)^2, you can apply the chain rule. Let's denote u = 2-x^4. So, (2-x^4)^2 becomes u^2. The chain rule states that if y = f(u), then dy/du * du/dx = dy/dx.
Differentiating u = 2-x^4 with respect to x, we get du/dx = -4x^3.
Thus, the first term (3x-1)^2 * d(2-x^4)^2/dx in dy/dx becomes:
(3x-1)^2 * 2 * (du/dx) = (3x-1)^2 * 2 * (-4x^3) = -8(3x-1)^2 * x^3
Similarly, differentiating (3x-1)^2 with respect to x, we get:
d(3x-1)^2/dx = 2(3x-1) * 3 = 6(3x-1)
Therefore, the second term (2-x^4)^2 * d(3x-1)^2/dx in dy/dx becomes:
(2-x^4)^2 * 6(3x-1) = 6(2-x^4)^2(3x-1)
Step 3: Combine the terms:
Now, we can combine the two terms we obtained from step 2:
dy/dx = -8(3x-1)^2 * x^3 + 6(2-x^4)^2(3x-1)
So, the derivative of the function y = (3x-1)^2(2-x^4)^2 is given by dy/dx = -8(3x-1)^2 * x^3 + 6(2-x^4)^2(3x-1).
Remember to simplify or factorize the expression further if necessary.