Introduction Of Derivatives Assessment Test

Explanation
The given expression is a product of two functions, (2+3x) and (1-x). To find the derivative, we can use the product rule. The product rule states that the derivative of a product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. Applying this rule, we get (2+3x)(-1) + (1-x)(3). Simplifying this expression gives -2-3x+3-3x, which further simplifies to -6x+1. Hence, the correct answer is 1-6x.

Explanation
To find the derivative of the function y=2x^2(x-1), we can use the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by u'(x)v(x) + u(x)v'(x). Applying this rule, we find that the derivative of y=2x^2(x-1) is y'=2(2x)(x-1) + 2x^2(1) = 4x(x-1) + 2x^2 = 4x^2 - 4x + 2x^2 = 6x^2 - 4x. Evaluating this derivative at x=-1, we get y'(-1) = 6(-1)^2 - 4(-1) = 6 - 4 = 2. Therefore, the correct answer is 2.

Explanation
The given question asks for the derivative of y with respect to x, which can be found using the power rule for differentiation. According to the power rule, when differentiating a function of the form (a + b)^n, the derivative is n(a + b)^(n-1) times the derivative of (a + b). In this case, y = (1 + x)^2, so the derivative of y with respect to x is 2(1 + x)^(2-1) times the derivative of (1 + x), which simplifies to 2(1 + x). Therefore, the correct answer is 2 + 2x.

Explanation
The given expression (2x+1) (3x+1) represents the product of two binomials. To find the derivative of this expression, we can use the product rule of differentiation. According to the product rule, the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. Applying this rule, we get the derivative of (2x+1) (3x+1) as (2)(3x+1) + (2x+1)(3) which simplifies to 6x+2 + 6x+3 = 12x+1.

Explanation
The correct answer is "Sec2 b". The derivative of sinb/cosb can be found using the quotient rule. The derivative of sinb is cosb, and the derivative of cosb is -sinb. Applying the quotient rule, we get (cosb*cosb - (-sinb*sinb)) / (cosb)^2, which simplifies to cos^2(b) + sin^2(b) / cos^2(b). Since sin^2(b) + cos^2(b) = 1, the derivative becomes 1 / cos^2(b), which is equivalent to Sec^2(b). Therefore, the correct answer is Sec^2(b).

Explanation
The given expression is a product of two functions: t^2 and sin(3t - 5). To find the derivative of this expression with respect to t, we can use the product rule. According to the product rule, the derivative of a product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. Applying the product rule to the given expression, we get 2t sin (3t - 5) + 3t^2 cos (3t - 5).

Explanation
The correct answer is -3x / √(2 – 3×2).

To find the derivative of the function √(2 – 3x^2), we can use the chain rule. The chain rule states that if we have a composite function, f(g(x)), then the derivative is given by f'(g(x)) * g'(x). In this case, the outer function is the square root and the inner function is 2 – 3x^2.

The derivative of the outer function, √u, is 1 / (2√u).

The derivative of the inner function, 2 – 3x^2, is -6x.

Using the chain rule, the derivative of the function √(2 – 3x^2) is (1 / (2√(2 – 3x^2))) * (-6x), which simplifies to -3x / √(2 – 3x^2).

Explanation
The given function is a polynomial function of degree 3. To find the value of the first derivative at x = 2, we need to differentiate the function. The derivative of x to the 3rd power is 3x^2, and the derivative of -6x is -6. The derivative of a constant term is 0. Therefore, the first derivative of f(x) is 3x^2 - 6. Plugging in x = 2 into the first derivative, we get f'(2) = 3(2)^2 - 6 = 12 - 6 = 6.

Explanation
The partial derivative of the function xy^2 - 5y + 6 with respect to x is given by taking the derivative of each term with respect to x while treating y as a constant. The derivative of xy^2 with respect to x is y^2, the derivative of -5y with respect to x is 0, and the derivative of 6 with respect to x is 0. Therefore, the partial derivative with respect to x is y^2 - 0 + 0 = y^2. Since the constant term -5 does not have an x term, it does not affect the derivative.

Explanation
The derivative of ln (cos x) can be found using the chain rule. The derivative of ln(u) is 1/u multiplied by the derivative of u. In this case, u = cos x, so the derivative of ln (cos x) is 1/(cos x) multiplied by the derivative of cos x. The derivative of cos x is -sin x. Therefore, the derivative of ln (cos x) is -sin x divided by cos x, which simplifies to -tan x.