Calculus I, II and III

College Algebra

To determine the first derivative of \( y \) with respect to \( x \) for each of the given functions, we will apply the rules of differentiation. The rules we will use include the power rule, the constant multiple rule, and the product rule. a. For the function \( y = 10 + 3x^2 + 2.5x^6 \), we apply the power rule and the constant multiple rule. The power rule states that the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \), and the constant multiple rule states that the derivative of a constant times a function is the constant times the derivative of the function. Taking the derivative term by term, we get: \[ \frac{dy}{dx} = \frac{d}{dx}(10) + \frac{d}{dx}(3x^2) + \frac{d}{dx}(2.5x^6) \] The derivative of a constant is zero, so the derivative of 10 is 0. Applying the power rule to the other terms, we have: \[ \frac{dy}{dx} = 0 + 3 \cdot 2x^{2-1} + 2.5 \cdot 6x^{6-1} \] Simplifying, we get: \[ \frac{dy}{dx} = 6x + 15x^5 \] b. For the function \( y = 2x(4x^2 + x - 3x^3) \), we will use the product rule in addition to the power rule. The product rule states that the derivative of a product of two functions \( u(x) \) and \( v(x) \) is \( u'(x)v(x) + u(x)v'(x) \). Let \( u(x) = 2x \) and \( v(x) = 4x^2 + x - 3x^3 \). Then, \( u'(x) = 2 \) and \( v'(x) = 8x + 1 - 9x^2 \) by applying the power rule to each term. Now, applying the product rule, we get: \[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \] Substituting \( u, u', v, \) and \( v' \), we have: \[ \frac{dy}{dx} = 2(4x^2 + x - 3x^3) + 2x(8x + 1 - 9x^2) \] Expanding the terms, we get: \[ \frac{dy}{dx} = 8x^2 + 2x - 6x^3 + 16x^2 + 2x - 18x^3 \] Combining like terms, we have: \[ \frac{dy}{dx} = -24x^3 + 24x^2 + 4x \] c. For the function \( y = 9x(100 - (4x)^2 - (3x)^2) \), we will again use the product rule. Let \( u(x) = 9x \) and \( v(x) = 100 - (4x)^2 - (3x)^2 \). Then, \( u'(x) = 9 \) and \( v'(x) = -2 \cdot 4x \cdot 2 - 2 \cdot 3x \cdot 2 \) by applying the power rule and the chain rule to each term. Now, applying the product rule, we get: \[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \] Substituting \( u, u', v, \) and \( v' \), we have: \[ \frac{dy}{dx} = 9(100 - (4x)^2 - (3x)^2) + 9x(-8x - 6x) \] Expanding the terms, we get: \[ \frac{dy}{dx} = 900 - 144x^2 - 81x^2 - 72x^2 - 54x^2 \] Combining like terms, we have: \[ \frac{dy}{dx} = 900 - 225x^2 - 72x^2 \] Simplifying further, we get: \[ \frac{dy}{dx} = 900 - 297x^2 \] These are the derivatives of the given functions without simplifying by combining terms.

Math

Determine the first derivative of y with respect to x for each of the following functions. No need to simplify the derivative by combining terms; just practice taking the derivative. a. \( y = 10 + 3x^2 + 2.5x^6 \) b. \( y = 2x(4x^2 + x - 3x^3) \) c. \( y = 9x(100 - (4x)^2 - (3x)^2) \)

Question

Show More

Solution

PrepMate

Ask a tutor

If you have any additional questions, you can ask one of our experts.