Given the function y=(2x^(3)-1)(3x^(2)-3+x), find (dy)/(dx) in any form. Answer: (dy)/(dx)=

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Given the function  y=(2x^(3)-1)(3x^(2)-3+x), find  (dy)/(dx) in any form. Answer:  (dy)/(dx)=

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Product Rule Explanation: To find the derivative of the given function, we will use the product rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Let's denote the first function as u(x) = 2x^3 - 1 and the second function as v(x) = 3x^2 - 3 + x.Derivative of u(x): First, we find the derivative of u(x) with respect to x. The derivative of 2x^3 is 6x^2, and the derivative of a constant (-1) is 0. So, the derivative of u(x), denoted as u'(x), is 6x^2.Derivative of v(x): Next, we find the derivative of v(x) with respect to x. The derivative of 3x^2 is 6x, the derivative of -3 is 0, and the derivative of x is 1. So, the derivative of v(x), denoted as v'(x), is 6x + 1.Apply Product Rule: Now, we apply the product rule. The derivative of y with respect to x, denoted as (dy)/(dx), is given by u'(x)v(x) + u(x)v'(x). Substituting the derivatives we found earlier, we get (dy)/(dx) = (6x^2)(3x^2 - 3 + x) + (2x^3 - 1)(6x + 1).Simplify Expression: We now need to simplify the expression. First, we distribute 6x^2 across the terms in the first part: (6x^2)(3x^2) + (6x^2)(-3) + (6x^2)(x). This simplifies to 18x^4 - 18x^2 + 6x^3.Combine Like Terms: Next, we distribute (2x^3 - 1) across the terms in the second part: (2x^3)(6x) + (2x^3)(1) - (1)(6x) - (1)(1). This simplifies to 12x^4 + 2x^3 - 6x - 1.Now, we combine like terms from both parts of the derivative. Adding 18x^4 to 12x^4 gives 30x^4. Adding 6x^3 to 2x^3 gives 8x^3. We already have -18x^2. And finally, we combine -6x and -1. So, the simplified form of the derivative is (dy)/(dx) = 30x^4 + 8x^3 - 18x^2 - 6x - 1.

Given the function $y=\left(2 x^{3}-1\right)\left(3 x^{2}-3+x\right)$ , find $\frac{d y}{d x}$ in any form. $\newline$ Answer: $\frac{d y}{d x}=$

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Given the function y=(2x3−1)(3x2−3+x) y=\left(2 x^{3}-1\right)\left(3 x^{2}-3+x\right) y=(2x3−1)(3x2−3+x), find dydx \frac{d y}{d x} dxdy​ in any form.\newlineAnswer: dydx= \frac{d y}{d x}= dxdy​=

Given the function $y=\left(2 x^{3}-1\right)\left(3 x^{2}-3+x\right)$ , find $\frac{d y}{d x}$ in any form. $\newline$ Answer: $\frac{d y}{d x}=$