AP Unit 3 Solutions

1. Identify outside u⁵ and inside u=2x³−1. Reason: the expression is a composition.
2. Outside differentiation gives 5(2x³−1)⁴; inside differentiation gives 6x². Reason: use the power rule at each layer.
3. Multiply to get 30x²(2x³−1)⁴. Reason: the chain rule combines the two rates.

Answer: f′(x)=30x²(2x³−1)⁴.

1. Identify u=4x²+1 and y=cos u. Reason: cosine is the outer function.
2. The outer derivative is −sin(4x²+1) and u′=8x. Reason: use the cosine and polynomial rules.
3. Multiply to get y′=−8x sin(4x²+1). Reason: the chain rule includes the inside rate.

Answer: y′=−8x sin(4x²+1).

1. The argument x²+3 is positive for every real x. Reason: x²≥0.
2. Use outside derivative 1/(x²+3) and inside derivative 2x. Reason: the logarithm is composed with a polynomial.
3. Multiply to get 2x/(x²+3). Reason: the chain rule applies throughout the real domain.

Answer: f′(x)=2x/(x²+3), all real x.

1. The layers are exponential, sine and 3x. Reason: each inside output feeds the next function.
2. Their derivative factors are e^(sin(3x)), cos(3x), and 3. Reason: evaluate each outer derivative at its inside input.
3. Multiply all three to get 3e^(sin(3x))cos(3x). Reason: repeated chain rules connect the layers.

Answer: f′(x)=3e^(sin(3x))cos(3x).

1. Use (f∘g)′(2)=f′(g(2))g′(2). Reason: the chain rule evaluates the outside rate at the inside output.
2. g(2)=4, so use f′(4)=5 rather than f′(2)=7. Reason: 4 is the input received by f.
3. Multiply 5(−3)=−15. Reason: combine the two derivative factors.

Answer: −15; f′(2)=7 is the unused distractor.

1. The height over time is H(V(t)). Reason: time changes volume, which changes height.
2. Use dH/dt=H′(8)V′(5)=0.25(−0.6). Reason: the chain rule combines the graph slope with the time rate.
3. The result is −0.15 m/min. Reason: liters cancel and the sign indicates decreasing height.

Answer: Height decreases at 0.15 m/min.

1. Differentiate the log layer and the exponential layer to obtain f′(x)=eˣ/(1+eˣ). Reason: the chain rule multiplies 1/(1+eˣ) by eˣ.
2. At 1 the exact value is e/(1+e)≈0.731059. Reason: evaluate only after differentiation.
3. The centered difference at 1 with h=0.001 is approximately 0.731058563, consistent with the analytic value. Reason: a numerical slope provides an independent local check, not a replacement for the derivation.

Answer: f′(x)=eˣ/(1+eˣ); f′(1)≈0.731059.

1. Verify 3²+4²=25, then differentiate to 2x+2y y′=0. Reason: the point lies on the curve and y depends on x.
2. Solve y′=−x/y, so y′(3,4)=−3/4. Reason: y=4 is nonzero.
3. Write y−4=−(3/4)(x−3). Reason: the derivative gives the tangent slope at the supplied point.

Answer: dy/dx=−x/y where y≠0; tangent y−4=−(3/4)(x−3).

1. Differentiate to 2x+y+x y′+2y y′=0. Reason: xy needs the product rule and y² needs the chain rule.
2. Collect terms: (x+2y)y′=−(2x+y). Reason: isolate the unknown local rate.
3. At (1,2), y′=−4/5. Reason: 1+2·2=5≠0 and the point satisfies the curve.

Answer: dy/dx=−4/5 at (1,2).

1. Differentiate eʸy′+2x=0. Reason: the exponential contains y(x).
2. Solve y′=−2x/eʸ; at (0,0) the slope is 0. Reason: eʸ is positive and the point lies on the curve.
3. Since eʸ=1−x²>0, the branch requires −1<x<1. Reason: a real logarithm y=ln(1−x²) needs a positive argument.

Answer: dy/dx=−2x/eʸ; slope 0 at (0,0); −1<x<1.

1. Substitute the claimed point: 1·0+sin 0=0. Reason: a tangent at a point requires curve membership.
2. Since 0≠1, (1,0) is not on the curve. Reason: the given equation excludes that point.
3. No tangent to this curve at that point is defined. Reason: differentiating and substituting cannot create a missing curve point.

Answer: No tangent slope at (1,0), because the point is not on the curve.

1. The point satisfies the circle. Reason: 3²+0²=9.
2. The y-as-a-function derivative formula would divide by 0. Reason: its finite-slope condition y≠0 fails.
3. The geometric tangent is vertical, x=3; there is no finite dy/dx there. Reason: the circle reaches its rightmost point and the limiting tangent direction is vertical.

Answer: Vertical tangent x=3; no finite dy/dx at that point.

1. Solve the quadratic in y: y=[−x+√(28−3x²)]/2 for the upper branch. Reason: the plus-square-root choice selects the larger y-value.
2. At x=1.2, y≈1.833105012. Reason: substitute without rounding intermediate values.
3. Use y′=−(2x+y)/(x+2y) to obtain approximately −0.869898. Reason: implicit differentiation gives the rate along this specific branch.

Answer: y≈1.833105012; tangent slope≈−0.869898.

1. The pair f(2)=5 gives f⁻¹(5)=2. Reason: the inverse swaps the input and output.
2. Use the inverse derivative at 5 as 1/f′(2). Reason: the corresponding original input is 2.
3. Compute 1/(−4)=−1/4. Reason: the supplied nonzero slope has a finite reciprocal.

Answer: f⁻¹(5)=2; (f⁻¹)′(5)=−1/4.

1. Solve y=3x+2 for x to get g(y)=(y−2)/3. Reason: inversion exchanges input and output.
2. Differentiate g to get g′(y)=1/3. Reason: it is a linear function.
3. At 8, g(8)=2 and 1/f′(2)=1/3. Reason: the original slope is 3 and the inverse identity predicts its reciprocal.

Answer: g(x)=(x−2)/3; g′(8)=1/3.

1. f(3)=10, so g(10)=3. Reason: the nonnegative domain selects the positive original input.
2. f′(3)=6, so g′(10)=1/6. Reason: the nonzero original slope has reciprocal inverse slope.
3. Differentiating √(x−1) gives 1/[2√(x−1)], which equals 1/6 at 10. Reason: the chain rule independently confirms the result.

Answer: g′(10)=1/6.

1. Solve a³+a=5 numerically to obtain a≈1.515980228. Reason: the inverse input 5 corresponds to the original input a.
2. f′(a)=3a²+1 and g′(5)=1/(3a²+1). Reason: the inverse derivative uses the corresponding original input.
3. Compute g′(5)≈0.126669. Reason: keep the unrounded root while evaluating the reciprocal; f′(5) uses the wrong original input.

Answer: a≈1.515980228; g′(5)≈0.126669.

1. f′(0)=0, so the reciprocal formula would require division by zero. Reason: its nonzero-slope hypothesis fails.
2. The inverse difference quotient at 0 is h^(1/3)/h=h^(−2/3). Reason: use the derivative definition directly.
3. This is unbounded positive as h→0, so no finite g′(0) exists. Reason: a derivative must be a finite limiting slope.

Answer: No finite g′(0); 1/0 is not a number.

1. The argument condition −1≤2x≤1 gives −1/2≤x≤1/2. Reason: arcsin has inputs in [−1,1].
2. Apply the inverse-sine and chain rules: y′=2/√(1−4x²). Reason: the inside rate is 2.
3. A finite derivative requires −1/2<x<1/2. Reason: the denominator is zero at the endpoints.

Answer: Function domain [−1/2,1/2]; y′=2/√(1−4x²) for −1/2<x<1/2.

1. Apply the inverse-cosine rule and inside rate 2x to get −2x/√(1−x⁴). Reason: the chain rule preserves the negative arccos rate.
2. At x=1/2 the value is −1/√(15/16)=−4/√15. Reason: evaluate the exact formula before simplifying.
3. The finite formula requires |x|<1. Reason: x²≤1 for the real function, with zero denominator at x=±1.

Answer: y′=−2x/√(1−x⁴) for |x|<1; y′(1/2)=−4/√15.

1. The inside derivative is 1/(2√x). Reason: the square-root power rule applies for x>0.
2. Use the inverse-tangent rule to get y′=1/[2√x(1+x)]. Reason: 1+(√x)²=1+x and the chain rule multiplies the inside rate.
3. At 0 the right quotient arctan(√h)/h is unbounded positive, since arctan u/u→1 as u→0. Reason: it is asymptotic to 1/√h.

Answer: y′=1/[2√x(1+x)] for x>0; no finite right derivative at 0.

1. Rewrite tan y=x. Reason: the stated branch makes tangent invertible.
2. Differentiate to sec²y·y′=1. Reason: y depends on x.
3. Use sec²y=1+tan²y=1+x², giving y′=1/(1+x²). Reason: the identity eliminates y and the denominator is positive.

Answer: d(arctan x)/dx=1/(1+x²).

1. (a) Differentiate cot y=x: −csc²y·y′=1, so y′=−1/(1+x²). Reason: csc²y=1+cot²y and the stated branch fixes the inverse.
2. (b) The arcsec rule is 1/(|x|√(x²−1)), so at −2 it is 1/(2√3). Reason: the branch sign requires |x| and |x|>1.
3. (c) The arccsc rule is −1/(|x|√(x²−1)), so at 2 it is −1/(2√3). Reason: the cosecant inverse has the negative rate and requires |x|>1.

Answer: (a) −1/(1+x²), all real x. (b) 1/(2√3). (c) −1/(2√3); secant/cosecant inverse formulas require |x|>1.

1. The outer structure is a product, so use f′=2x sin(3x)+x²·d[sin(3x)]/dx. Reason: both factors vary.
2. The sine factor is a composition with derivative 3cos(3x). Reason: the chain rule includes the inside rate 3.
3. Combine to get 2x sin(3x)+3x²cos(3x). Reason: all product and chain contributions are present.

Answer: f′(x)=2x sin(3x)+3x²cos(3x); power, product, sine and chain rules.

1. Rewrite y=ln(x+1)−ln(x−1). Reason: both factors are positive on x>1, so the logarithm quotient identity is valid.
2. Differentiate to 1/(x+1)−1/(x−1). Reason: use the log, difference and chain rules.
3. Combine to −2/(x²−1), retaining x>1. Reason: a common denominator simplifies the formula without changing the stated domain.

Answer: y′=−2/(x²−1), x>1.

1. The outer structure is a quotient with denominator 1+x≠0. Reason: the expression is defined only for x≠−1.
2. Differentiate the numerator by the chain rule to get 2x e^(x²). Reason: x² is the exponential input.
3. Use the quotient rule: [2x e^(x²)(1+x)−e^(x²)]/(1+x)²=e^(x²)(2x²+2x−1)/(1+x)². Reason: numerator and denominator both vary.

Answer: f′(x)=e^(x²)(2x²+2x−1)/(1+x)², x≠−1.

1. f′=5x⁴−6x². Reason: use polynomial rules and remove the constant rate.
2. f″=20x³−12x and f‴=60x²−12. Reason: repeatedly differentiate the preceding derivative.
3. f‴(1)=48. Reason: evaluate the third derivative, not the original function or a cube of it.

Answer: f′=5x⁴−6x²; f″=20x³−12x; f‴=60x²−12; f‴(1)=48.

1. Velocity is s′(t)=2cos(2t). Reason: differentiate position with the chain rule.
2. Acceleration is s″(t)=−4sin(2t). Reason: differentiate velocity and include the inside rate again.
3. At π/4, acceleration is −4 m/s². Reason: sin(π/2)=1 and a second derivative has output per time squared units.

Answer: Velocity 2cos(2t) m/s; acceleration −4sin(2t) m/s²; at π/4 it is −4 m/s².

1. First f′=eˣ(sin x+cos x). Reason: the product rule accounts for both factors.
2. Differentiate again to f″=eˣ(sin x+cos x)+eˣ(cos x−sin x)=2eˣcos x. Reason: combine the two product-rule terms.
3. Differentiate that product to f‴=2eˣ(cos x−sin x). Reason: the derivative of cosine introduces its minus sign.

Answer: f″=2eˣcos x; f‴=2eˣ(cos x−sin x).

1. First 2x+2y y′=0, so y′=−x/y. Reason: implicit differentiation treats y as depending on x.
2. Differentiate the first equation again: 2+2(y′)²+2y y″=0. Reason: differentiating y y′ requires the product rule.
3. Thus y″=−[1+(y′)²]/y; at (3,4), y′=−3/4, giving y″=−(25/16)/4=−25/64. Reason: the branch has y=4≠0, so division and substitution are valid.

Answer: y″(3,4)=−25/64.