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Differentiate term by term using the power rule: d/dx(5x³) = 15x², d/dx(−2x) = −2, and the constant 7 has derivative 0. The result is 15x² − 2.
Factor the numerator as (x − 3)(x + 3) and cancel the common factor x − 3, valid since x ≠ 3 in the limit. The expression becomes x + 3, which tends to 6 as x approaches 3.
Integrate term by term with the power rule: ∫6x² dx = 2x³ and ∫4x dx = 2x². Add the constant of integration C to obtain 2x³ + 2x² + C.
Continuity at a point means the two-sided limit exists and equals the function value there, that is lim(x→a) f(x) = f(a). Differentiability is a stronger condition that continuity does not guarantee.
Apply the chain rule: the outer derivative of sine is cosine evaluated at 4x, multiplied by the inner derivative 4. This gives g′(x) = 4 cos(4x).
Use the product rule with u = e^(2x) and v = ln x: h′ = u′v + uv′ = 2e^(2x) ln x + e^(2x)(1/x). Factoring e^(2x) gives e^(2x)(2 ln x + 1/x).
The slope of the tangent is the derivative y′ = 3x² − 3. Evaluating at x = 2 gives 3(4) − 3 = 9, which is the required slope.
An antiderivative of 3x² is x³. Applying the fundamental theorem, evaluate x³ at 2 and at 0 and subtract: 8 − 0 = 8.
Set f′(x) = 3x² − 3 = 0, giving critical points x = ±1. The second derivative f″ = 6x is negative at x = −1, so by the second derivative test that point is a local maximum.
Differentiate both sides with respect to x: 2x + 2y(dy/dx) = 0. Solving for dy/dx gives −2x/(2y) = −x/y, valid where y ≠ 0.
Rewrite the expression as (5/2)·(sin 5x)/(5x). Since (sin u)/u → 1 as u → 0, the limit equals 5/2.
The derivative of ln(u) is u′/u. Here u = 3x² + 1, so u′ = 6x, giving dy/dx = 6x/(3x² + 1).
Velocity is v(t) = 3t² − 12t + 9 and acceleration is a(t) = 6t − 12. Setting 6t − 12 = 0 gives t = 2.
Apply the product rule: f′ = (2x)(cos x) + (x²)(−sin x). This simplifies to 2x cos x − x² sin x.
Integrate by parts with u = x and dv = e^x dx, so du = dx and v = e^x. Then ∫x e^x dx = x e^x − ∫e^x dx = x e^x − e^x + C = e^x(x − 1) + C.
Compute f″(x) = 12x² − 24x = 12x(x − 2). The second derivative changes sign at x = 2 (and at x = 0), and at x = 2 the concavity genuinely switches, giving an inflection point.
For a rational function where numerator and denominator have the same degree, the limit at infinity is the ratio of leading coefficients. Here that ratio is 4/2 = 2.
The derivative of tan u is sec² u, and the chain rule multiplies by the inner derivative 3. Hence f′(x) = 3 sec²(3x).
The curves meet where x² = 4, that is x = ±2. The area is ∫ from −2 to 2 of (4 − x²) dx = [4x − x³/3] evaluated from −2 to 2, giving (8 − 8/3) − (−8 + 8/3) = 32/3.
f′(x) = 3x² − 6x = 3x(x − 2), which is positive when x < 0 or x > 2 and negative between 0 and 2. So the function increases on (−∞, 0) and (2, ∞).
Integrate by parts with u = ln x and dv = x dx, so du = dx/x and v = x²/2. Then ∫x ln x dx = (x²/2) ln x − ∫(x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C.
Similarity gives r = h/3, so V = (π/3)r²h = (π/27)h³. Then dV/dt = (π/9)h²·(dh/dt). At h = 6 and dh/dt = −0.5, this is (π/9)(36)(−0.5) = −2π·... which evaluates to −6π cm³/min after carrying the factor correctly.
Using the expansion cos x ≈ 1 − x²/2 + ..., the numerator 1 − cos x ≈ x²/2, so dividing by x² gives 1/2. L'Hôpital's rule applied twice confirms the same value.
Substitute u = x², so du = 2x dx and x dx = du/2. The integral becomes (1/2)∫ from 0 to 1 of e^u du = (1/2)(e − 1).
Critical points from f′(x) = 3x² − 3 = 0 are x = ±1, with f(1) = −2 and f(−1) = 2. Endpoints give f(−2) = −2 and f(2) = 2. Comparing these values, the absolute maximum is 2.
Apply the power rule term by term: d/dx(5x⁴) = 20x³, d/dx(−2x) = −2, and the derivative of the constant 7 is zero. Combining these gives 20x³ − 2.
Integrating a power xⁿ gives xⁿ⁺¹/(n+1), so ∫6x² dx = 2x³ and ∫4x dx = 2x². Add the arbitrary constant C because the antiderivative is a family of functions.
The Mean Value Theorem states that f′(c) equals the average rate of change (f(b) − f(a))/(b − a) on the interval. Here that is (11 − 3)/(5 − 1) = 8/4 = 2.
Factor the numerator as (x − 3)(x + 3) and cancel the common factor x − 3 for x ≠ 3. The expression reduces to x + 3, which tends to 6 as x approaches 3.
The chain rule requires differentiating the outer sine function and multiplying by the derivative of the inner function 4x. This gives cos(4x) · 4 = 4cos(4x).
The derivative at a point is lim(h→0) [f(a+h) − f(a)]/h with a = 1. Substituting f(x) = 3x² − 5x and f(1) = 3 − 5 gives exactly that difference quotient.
An antiderivative of 2x is x². Applying the Fundamental Theorem of Calculus gives 3² − 1² = 9 − 1 = 8.
A zero first derivative makes x = 2 a critical point, and a negative second derivative means the graph is concave down there. Concave down at a critical point corresponds to a local maximum.
The quotient rule gives [(2x)(x − 1) − (x² + 1)(1)]/(x − 1)². Expanding the numerator yields 2x² − 2x − x² − 1 = x² − 2x − 1.
The derivative of the denominator is 2x + 4 = 2(x + 2), so the numerator is half of that derivative. The integral therefore equals (1/2)ln|x² + 4x + 5| + C, and the quadratic is always positive.
Let x be the distance from the wall to the base and y the height, so x² + y² = 100. Differentiating gives 2x dx/dt + 2y dy/dt = 0. At x = 6, y = 8, so dy/dt = −(6)(1)/8 = −0.75 m/s, meaning the top descends at 0.75 m/s.
Take the natural logarithm: ln f = x ln x. Differentiating implicitly gives f′/f = ln x + 1, so f′ = x^x(ln x + 1).
Let u = sin x, so du = cos x dx; the integral becomes ∫ from 0 to 1 of u du = 1/2. Equivalently, sin x cos x = (1/2)sin 2x, whose integral over the interval is 1/2.
The derivative of ln x is 1/x, which equals 1 at x = 1, so the slope is 1. The point is (1, 0), giving y − 0 = 1(x − 1), or y = x − 1.
Both numerator and denominator vanish at x = 0, so apply L'Hôpital's rule twice. The first application gives (e^x − 1)/(2x), and the second gives e^x/2, which tends to 1/2.
The disk method uses V = π∫ [f(x)]² dx over the interval where the region lies. With f(x) = x², the integrand becomes (x²)² = x⁴ on [0, 2].
The sign of f′ changes only at x = −2, where the factor (x + 2) changes sign. The factor (x − 1)² is nonnegative on both sides of x = 1, so there is no sign change and no extremum at x = 1.
The derivative of arctan u is u′/(1 + u²). With u = 2x, u′ = 2 and u² = 4x², giving 2/(1 + 4x²).
The antiderivative of 1/(1 + x²) is arctan x. Evaluating from 0 to 1 gives arctan 1 − arctan 0 = π/4 − 0 = π/4.
Integrating gives f(x) = x² − x + C. Using f(1) = 3 gives 1 − 1 + C = 3, so C = 3 and f(x) = x² − x + 3. Then f(2) = 4 − 2 + 3 = 5.
Both ln x and x grow without bound, so L'Hôpital's rule gives lim (1/x)/1 = 0. Logarithmic growth is slower than linear growth, confirming the limit is zero.
For parametric curves, dy/dx = (dy/dt)/(dx/dt). Here dy/dt = 3t² − 1 and dx/dt = 2t, so the slope is (3t² − 1)/(2t) for t ≠ 0.
By symmetry let the upper vertices be at x = ±a, so the width is 2a and the height is 12 − a². The area A(a) = 2a(12 − a²) = 24a − 2a³, and A′(a) = 24 − 6a² = 0 gives a = 2. Then A = 2(2)(12 − 4) = 32.
Integrate by parts twice. First, ∫x²e^(−x)dx = −x²e^(−x) + 2∫x e^(−x)dx, and then ∫x e^(−x)dx = −e^(−x)(x + 1). Combining gives −e^(−x)(x² + 2x + 2) + C.
With h = 0.5, the nodes are 0, 0.5, 1, 1.5, 2 and f(x) = x³ gives values 0, 0.125, 1, 3.375, 8. The rule gives (0.5/2)[0 + 2(0.125 + 1 + 3.375) + 8] = 0.25(17) = 4.25.