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Differentiate $y-(2x+1)^{5}(x^{3}-x+1)^{4}$

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Evaluate integers raised to rational exponents

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Q. Differentiate

y-(2x+1)^{5}(x^{3}-x+1)^{4}

Define Functions: Let's denote the first function as $u = (2x+1)^{5}$ and the second function as $v = (x^{3}-x+1)^{4}$ . We will need to find the derivatives $u'$ and $v'$ of these functions.
Find Derivatives: First, we find the derivative of $u = (2x+1)^{5}$ . Using the chain rule, we get $u' = 5\cdot(2x+1)^{4}\cdot2$ , since the derivative of the inner function $2x+1$ is $2$ .
Apply Product Rule: Now, we find the derivative of $v = (x^{3}-x+1)^{4}$ . Again, using the chain rule, we get $v' = 4\cdot(x^{3}-x+1)^{3}\cdot(3x^{2}-1)$ , since the derivative of the inner function $x^{3}-x$ is $3x^{2}-1$ .
Simplify Expression: Using the product rule, the derivative of $y = uv$ is $y' = u'v + uv'$ . Substituting the derivatives we found, we get $y' = (5(2x+1)^{4}2)(x^{3}-x+1)^{4} + (2x+1)^{5}(4(x^{3}-x+1)^{3}(3x^{2}-1))$ .
Final Derivative: Now we simplify the expression for $y'$ . We distribute the terms and combine like terms if possible. $y' = 10\cdot(2x+1)^{4}\cdot(x^{3}-x+1)^{4} + (2x+1)^{5}\cdot4\cdot(x^{3}-x+1)^{3}\cdot(3x^{2}-1)$ .
Final Derivative: Now we simplify the expression for $y'$ . We distribute the terms and combine like terms if possible. $y' = 10\cdot(2x+1)^{4}\cdot(x^{3}-x+1)^{4} + (2x+1)^{5}\cdot4\cdot(x^{3}-x+1)^{3}\cdot(3x^{2}-1)$ .The expression for $y'$ is now fully simplified, and we cannot combine terms further because they are not like terms. Therefore, the derivative of the function $y = (2x+1)^{5}(x^{3}-x+1)^{4}$ is: $y' = 10\cdot(2x+1)^{4}\cdot(x^{3}-x+1)^{4} + 12\cdot(2x+1)^{5}\cdot(x^{3}-x+1)^{3}\cdot(x^{2}-\frac{1}{3})$ .

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