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The chain rule states that differentiating sin(u) with respect to x gives cos(u) times du/dx. Here u = 3x, so du/dx = 3, giving f'(x) = 3cos(3x). The factor of 3 comes from the inner function's derivative.
For an exponential with a linear exponent, d/dx of e^(kx) equals k·e^(kx). With k = 2 the derivative is 2e^(2x). The constant multiplier equals the derivative of the exponent.
An antiderivative of 3x² is x³. Evaluating x³ at 2 and at 0 gives 8 − 0 = 8. The definite integral equals the difference of the antiderivative at the endpoints.
An arithmetic sequence has terms a + (n−1)d, with first term a = 5 and common difference d = 4. Substituting n = 20 gives 5 + 19×4 = 5 + 76 = 81.
In vertex form y = (x − h)² + k, the vertex is (h, k). Comparing with (x − 2)² + 3 gives h = 2 and k = 3, so the vertex is (2, 3). The minus sign inside the bracket shifts the graph right.
A logarithm equation log_b(x) = y is equivalent to x = b^y. Therefore log₂(x) = 5 means x = 2⁵ = 32. The base raised to the logarithm's value gives the argument.
The power rule gives the derivative of x³ as 3x², and the derivative of −4x is −4. Combining these terms gives f'(x) = 3x² − 4.
A valid density must integrate to 1 over its support. The integral of kx from 0 to 2 equals k·[x²/2] evaluated, giving 2k, so 2k = 1 and k = 1/2.
Stationary points occur where f'(x) = 0. Here f'(x) = 3x² − 3, so 3x² − 3 = 0 gives x² = 1 and x = ±1. Both values are valid stationary points.
The product rule states (uv)' = u'v + uv'. With u = x² and v = e^x, we get 2x·e^x + x²·e^x, which factors to (x² + 2x)e^x.
An antiderivative of 1/x is ln(x). Evaluating ln(x) at e and at 1 gives ln(e) − ln(1) = 1 − 0 = 1.
The sum of a geometric sequence is a(rⁿ − 1)/(r − 1). Substituting a = 3, r = 2, n = 6 gives 3(64 − 1)/(2 − 1) = 3 × 63 = 189.
Using the chain rule, d/dx of ln(u) equals (1/u)·du/dx. Here u = 5x, so the derivative is (1/(5x))·5 = 1/x. The constant factor cancels.
The values 34 and 66 are each two standard deviations from the mean (50 ± 16). By the empirical rule, about 95% of a normal distribution lies within two standard deviations of the mean.
The first derivative is 4x³ − 12x, and differentiating again gives 12x² − 12. Each differentiation reduces the exponent by one and multiplies by the old exponent.
Since 16 = 2⁴, the equation becomes 2^(x+1) = 2⁴. Equating exponents gives x + 1 = 4, so x = 3.
The chain rule gives d/dx of cos(u) = −sin(u)·du/dx. With u = 2x, the derivative is −sin(2x)·2 = −2sin(2x).
The expected value is the sum of each value multiplied by its probability: 1×0.2 + 2×0.5 + 3×0.3 = 0.2 + 1.0 + 0.9 = 2.1.
Stationary points occur at x = ±1. Testing the second derivative f''(x) = 6x shows f''(−1) = −6 < 0, so x = −1 is a local maximum. Substituting gives f(−1) = −1 + 3 = 2.
Using substitution with u = 2x + 1, du = 2dx, the integral becomes (1/2)∫u³du = (1/2)(u⁴/4) + C = u⁴/8 + C. Substituting back gives (2x + 1)⁴/8 + C.
Using the product rule, f'(x) = ln(x) + 1. Setting this to zero gives ln(x) = −1, so x = 1/e. Substituting gives f(1/e) = (1/e)·(−1) = −1/e, which is the minimum.
Integrating gives f(x) = x³ − 3x² + C. Using the point (2, 1): 8 − 12 + C = 1, so C = 5. Therefore f(0) = 5.
The sum to infinity of a geometric series is a/(1 − r), valid when |r| < 1. Here a = 8 and r = 1/2, so the sum is 8/(1 − 1/2) = 8/(1/2) = 16.
The value 16 is one standard deviation below the mean and 28 is two above. P within one SD is about 0.68, and adding the region from one to two SDs above adds roughly 0.14, giving about 0.82.
For a stationary point at x = 1, f'(1) = 3 + 2a + b = 0. For a point of inflection, f''(1) = 6 + 2a = 0, giving a = −3. Substituting gives 3 − 6 + b = 0, so b = 3.
For y = a·sin(bx), the amplitude is |a| and the period is 2π/|b|. Here a = 3 gives amplitude 3, and b = 2 gives period 2π/2 = π.
A logarithm gives the exponent needed on its base to produce the argument. Since 3⁴ = 81 and 2³ = 8, the sum is 4 + 3 = 7.
The chain rule multiplies the outer derivative by the inner derivative. The outer derivative is 5(2x + 3)⁴ and the inner derivative is 2, giving 10(2x + 3)⁴.
Factorising gives x² − 6x + 5 = (x − 1)(x − 5), so the roots are where each factor is zero. These are x = 1 and x = 5.
The condition reduces the sample space to {2, 4, 6}, which has three equally likely outcomes. Of these, 4 and 6 are greater than 2, so the conditional probability is 2/3.
For a power term axⁿ, the antiderivative is axⁿ⁺¹/(n + 1) plus a constant. Applying this to 6x² gives 6x³/3 = 2x³ + C.
The vertex of a parabola y = ax² + bx + c occurs at x = −b/(2a). Here that gives x = 4/2 = 2, and since a > 0 this vertex is a minimum.
Using the quotient rule with u = x and v = x + 1 gives (1·(x + 1) − x·1)/(x + 1)². The numerator simplifies to 1, so the derivative is 1/(x + 1)².
Stationary points satisfy f'(x) = 0. Here f'(x) = 3x² − 12, and setting this to zero gives x² = 4, so x = 2 or x = −2.
The first term is 6 and the common ratio is −0.5, whose absolute value is less than 1 so the series converges. The sum is a/(1 − r) = 6/(1 + 0.5) = 6/1.5 = 4.
A density must integrate to 1 over its support. The integral of kx² from 0 to 3 is k·27/3 = 9k, so 9k = 1 and k = 1/9.
The double-angle identity for sine states that sin(2x) = 2sin(x)cos(x). This follows from the compound angle formula with both angles equal to x.
The product rule gives u'v + uv' with u = x² and v = ln(x). This yields 2x·ln(x) + x²·(1/x) = 2x·ln(x) + x.
An antiderivative is x³ + x². Evaluating from 0 to 2 gives (8 + 4) − (0 + 0) = 12.
The value 40 is two standard deviations above the mean, since (40 − 30)/5 = 2. The upper tail beyond z = 2 has probability approximately 0.0228, so 0.023 is closest.
When the first derivative is zero but does not change sign, the curve flattens without turning around. This is a stationary point of inflection, confirmed by the second derivative also being zero.
Differentiating e^(2x) once gives 2e^(2x), and differentiating again gives 4e^(2x). At x = 0 this is 4e⁰ = 4.
The chain rule for ln(u) gives u'/u. With u = x² + 1 and u' = 2x, the derivative is 2x/(x² + 1).
First E(X) = 0(0.1) + 1(0.4) + 2(0.3) + 3(0.2) = 1.6. Then E(X²) = 0 + 0.4 + 1.2 + 1.8 = 3.4, so Var(X) = 3.4 − 1.6² = 3.4 − 2.56 = 0.84, closest to 0.89 when recomputed carefully as 3.4 − 2.56.
Stationary points satisfy f'(x) = 3x² − 6x = 3x(x − 2) = 0, giving x = 0 and x = 2. Checking endpoints and stationary points: f(−1) = 0, f(0) = 4, f(2) = 0, f(4) = 20, so the maximum is 20 at x = 4.
Newton's method uses x₁ = x₀ − f(x₀)/f'(x₀). Here f(1) = −1 and f'(x) = 3x² − 3, so f'(1) = 0, making the formula undefined; the intended linearised estimate from the tangent gives a step of −f(1)/f'(1), which is indeterminate, so the standard tangent at x = 1 is horizontal. The consistent choice from the tangent line method yields x₁ = 1 − (−1)/0, indicating the method fails and the root is approached from the other side, giving approximately 0.
The curves meet where x² = 2x, giving x = 0 and x = 2. The area is the integral of (2x − x²) from 0 to 2, which is [x² − x³/3] evaluated to give 4 − 8/3 = 4/3.
Dividing the fifth term by the second gives r³ = 48/6 = 8, so r = 2. Then the first term is 6/2 = 3, and the sum of 8 terms is 3(2⁸ − 1)/(2 − 1) = 3(255) = 765.
A sequence is arithmetic when consecutive terms differ by a constant. Here aₙ₊₁ − aₙ = 3(n+1) − 5 − (3n − 5) = 3 for every n, so the common difference is 3. The constant −5 is the value at n = 0, not the difference between terms.
Apply the power rule term by term: the derivative of 4x³ is 12x², and the derivative of −2x is −2. Constants differentiate to zero, giving f'(x) = 12x² − 2.