Calculating Higher-Order Derivatives - AP Calculus AB

For $f(x)=\sin(x)+\cos(x)$ , what is the third derivative $f^{(3)}(x)$ ?
Answer: $f^{(3)}(x)= sin(x)-Ìos(x)$ Sums of sine and cosine have additive derivatives. The first is f'(x) = cos(x) - sin(x). The second is -sin(x) - cos(x), and third is sin(x) - cos(x). A common stopping error is not reaching the third. Patterns cycle every four derivatives. A transferable strategy is to differentiate term by term, observing cyclic behavior for trig functions.
If $f(x)=(4x-1)^5$ , what is the second derivative $f''(x)$ ?
Answer: $f''(x)=320(4x-1)^3$ Chain rule for powers leads to higher-order derivatives with reducing exponents. The first is f'(x) = 20 $(4x - 1)^4$ . The second is f''(x) = 320 $(4x - 1)^3$ , multiplying coefficients carefully. A common stopping error is halting after one differentiation. Verify by expanding if needed. A transferable strategy is to apply the chain rule repeatedly, updating the power and coefficient each time.
Let $q(x)=\sqrt{x}=x^{1/2}$ for $x>0$ . What is $q''(x)$ ?
Answer: $-\dfrac{1}{4x^{3/2}}$ For q(x) = √x = $x^{1/2}$ , we use the power rule with fractional exponents. First, q'(x) = (1/2) $x^{-1/2}$ = 1/(2√x). For the second derivative, q''(x) = (1/2)(-1/2) $x^{-3/2}$ = -1/4 · $x^{-3/2}$ = -1/( $4x^{3/2}$ ). Notice how the exponent decreases by 1 each time: 1/2 → -1/2 → -3/2. A common mistake is incorrectly handling fractional exponents or forgetting the negative sign that appears in the second derivative. When differentiating functions with fractional exponents, carefully track how the exponent changes and remember that $x^{-n/m}$ = 1/ $x^{n/m}$ for the final answer.
Let $f(x)=(6x^3-2x+9)$ . What is the second derivative $f''(x)$ ?
Answer: $f''(x)=36x$ Polynomials like ( f(x) = $6x^3$ - 2x + 9 ) simplify in higher-order derivatives as degrees decrease. First: ( f'(x) = $18x^2$ - 2 ), second: ( f''(x) = 36x ), confirming A. Successive power rule application eliminates constants and lowers powers. Common error: including constant terms in derivatives, but they vanish after first. Always differentiate term by term without skipping. Strategy: note the derivative order equals how many times powers reduce, helping predict final form.
If $f(x) = \sin(x^2)$ , what is the second derivative $f''(x)$ ?
Answer: $f''(x)=2\cos(x^2)-4x^2\sin(x^2)$ For composite functions like $f(x) = \sin(x^2)$ , higher-order derivatives use the chain rule multiple times. The first derivative is $f'(x) = \cos(x^2) \cdot 2x$ , and the second requires the product rule: $f''(x) = -\sin(x^2) \cdot 2x \cdot 2x + \cos(x^2) \cdot 2 = 2\cos(x^2) - 4x^2 \sin(x^2)$ , matching choice A. Successive application combines chain and product rules carefully. A common error is stopping after the first chain rule application and ignoring the product in the second derivative. Double-check by expanding terms fully before simplifying. Always identify inner and outer functions to apply rules systematically for any order.
For $y=(2x+1)e^{x}$ , what is the second derivative $y''$ ?
Answer: $y''=e^x(2x+5)$ Calculating higher-order derivatives requires repeated application of differentiation rules, such as the product rule for ( y = (2x + 1) $e^x$ ). The first derivative is ( y' = $2e^x$ + (2x + 1) $e^x$ = $e^x$ (2x + 3) ), and the second is ( y'' = $e^x$ (2) + $e^x$ (2x + 3) = $e^x$ (2x + 5) ), confirming choice A. Each step combines the derivative of the exponential and the linear factor. A frequent mistake is forgetting to apply the product rule twice, leading to incomplete terms like just ( $2e^x$ ). Ensure you differentiate every part successively without omitting factors. A useful strategy is to factor out common terms like ( $e^x$ ) to simplify verification at each order.
Let $u(x)=x^2\ln x$ for $x>0$ . What is $u''(x)$ ?
Answer: $2\ln x+3$ For u(x) = x²ln x where x > 0, we need the product rule twice. First, u'(x) = 2x ln x + x²(1/x) = 2x ln x + x. For the second derivative, differentiate each term: d/dx(2x ln x) = 2ln x + 2x(1/x) = 2ln x + 2, and d/dx(x) = 1. Therefore, u''(x) = 2ln x + 2 + 1 = 2ln x + 3. A common error is forgetting that d/dx(ln x) = 1/x when applying the product rule to the first term. Always carefully track each application of differentiation rules when products involve logarithms.
For $p(x)=\dfrac{1}{x^2}$ with $x\ne0$ , what is $p''(x)$ ?
Answer: $\dfrac{6}{x^4}$ To find p''(x) for p(x) = 1/x² = x⁻², we rewrite using negative exponents and apply the power rule twice. First, p'(x) = -2x⁻³ = -2/x³. Then, p''(x) = -2(-3)x⁻⁴ = 6x⁻⁴ = 6/x⁴. The key is recognizing that differentiating x⁻ⁿ gives -nx⁻ⁿ⁻¹, so negative exponents become more negative with each differentiation. A common error is mishandling the signs or incorrectly applying the power rule to negative exponents. When working with rational functions like 1/xⁿ, converting to negative exponents (x⁻ⁿ) makes differentiation more straightforward and helps avoid quotient rule complications.
Let $s(x)=\cos(3x)$ . What is $s''(x)$ ?
Answer: $-9\cos(3x)$ To find s''(x) for s(x) = cos(3x), we apply the chain rule twice. First, s'(x) = -sin(3x)·3 = -3sin(3x). Then s''(x) = -3cos(3x)·3 = -9cos(3x). The pattern emerges: each differentiation introduces a factor of 3 (from the chain rule) and cycles through cos → -sin → -cos → sin. A common mistake is forgetting to multiply by the derivative of the inner function (3) at each step. For trig functions with linear arguments, remember that the nth derivative introduces a factor of the argument's coefficient raised to the nth power.
Let $y=xe^{-x}$ . What is the second derivative $y''$ ?
Answer: $y''=e^{-x}(x-2)$ Product of x and exponential uses product rule. The first is y' = $e^{-x}$ (1 - x). The second is y'' = $e^{-x}$ (x - 2), factoring properly. A common error is stopping at first or sign errors in exponential. Each differentiation involves both product and chain. For such products, apply rules successively, factoring the exponential out.
Let $q(x)=e^{2x}$ . What is $q^{(3)}(x)$ ?
Answer: $8e^{2x}$ For q(x) = $e^{2x}$ , we need the third derivative q'''(x). Using the chain rule repeatedly: q'(x) = $2e^{2x}$ , q''(x) = $4e^{2x}$ , and q'''(x) = $8e^{2x}$ . Each differentiation multiplies by the derivative of the exponent (which is 2), giving us powers of 2: 2¹, 2², 2³. A common mistake is forgetting to apply the chain rule consistently or losing track of the coefficient. For exponential functions with linear exponents, the nth derivative of $e^{kx}$ is $k^n$ · $e^{kx}$ .
Let $g(x)=x^3\sin x$ . What is $g''(x)$ ?
Answer: $6x\sin x+6x^2\cos x-x^3\sin x$ Finding g''(x) for g(x) = x³sin x requires the product rule twice. First, g'(x) = 3x²sin x + x³cos x (using (uv)' = u'v + uv'). For the second derivative, we apply the product rule to each term: d/dx(3x²sin x) = 6x sin x + 3x²cos x and d/dx(x³cos x) = 3x²cos x - x³sin x. Combining these gives g''(x) = 6x sin x + 6x²cos x - x³sin x. A common mistake is forgetting to apply the product rule to both terms in g'(x). Remember: when differentiating products of functions multiple times, carefully track each application of the product rule.
A particle's position is $s(t)=t^4-6t^2+3t$ . What is the value of $s''(t)$ ?
Answer: $12t^2-12$ To find the second derivative s''(t) of s(t) = t⁴ - 6t² + 3t, we must differentiate twice. First, s'(t) = 4t³ - 12t + 3 using the power rule on each term. Then, differentiating again gives s''(t) = 12t² - 12 + 0 = 12t² - 12. A common error is stopping after the first derivative or incorrectly applying the power rule during the second differentiation. Notice how the constant term 3t becomes 3 after the first derivative, then 0 after the second. When finding higher-order derivatives, systematically apply differentiation rules to each term, tracking how powers decrease with each application.
For $y=x^2 ln(x)$ with $x>0$ , what is the second derivative $y''$ ?
Answer: $y''=2 ln(x)+3$ Products of polynomials and logs require product rule for derivatives. The first is y' = 2x ln(x) + x. The second is y'' = 2 ln(x) + 3, simplifying terms. A common error is stopping at the first or forgetting the 1/x from ln. Domain x > 0 is key. For such functions, use product rule successively, combining like terms.
For $y=\frac{1}{x^2}+7$ , what is the third derivative $y^{(3)}(x)$ ?
Answer: $y^{(3)}(x)=-\frac{24}{x^5}$ Rational functions' higher derivatives involve increasing negative exponents. The first is y' = -2/ $x^3$ . The second is 6/ $x^4$ , and third is -24/ $x^5$ . A common error is stopping at second or sign errors. Constants like 7 vanish early. For inverses, use power rule repeatedly, alternating signs with each differentiation.
A model uses $h(x)=e^{3x}$ . What is the value of the third derivative $h^{(3)}(x)$ ?
Answer: $27e^{3x}$ For h(x) = e³ˣ, finding the third derivative requires applying the chain rule three times. First, h'(x) = e³ˣ · 3 = 3e³ˣ. Second, h''(x) = 3e³ˣ · 3 = 9e³ˣ. Third, h'''(x) = 9e³ˣ · 3 = 27e³ˣ. Notice the pattern: each differentiation multiplies by 3 (the coefficient in the exponent), so the nth derivative is 3ⁿe³ˣ. A common mistake is forgetting to apply the chain rule at each step or losing track of the accumulating coefficients. For exponential functions with linear exponents like $e^{ax}$ , the nth derivative follows the pattern aⁿ $e^{ax}$ , making higher-order derivatives predictable.
Let $F(x)=\ln(x^2)$ for $x\ne0$ . What is the value of $F''(x)$ ?
Answer: $-\dfrac{2}{x^2}$ For F(x) = ln(x²), we can use the chain rule or first simplify using logarithm properties: ln(x²) = 2ln|x|. Taking the simpler approach, F'(x) = 2(1/x) = 2/x for x ≠ 0. Then F''(x) = 2(-1/x²) = -2/x². Alternatively, using the chain rule directly: F'(x) = (1/x²)(2x) = 2/x, and F''(x) = -2/x². A common error is forgetting the negative sign when differentiating 1/x or misapplying the chain rule. When dealing with logarithmic functions, simplifying using log properties before differentiating often makes the calculation clearer and reduces errors.
If $f(x)=(2x^3-1)^2$ , what is the second derivative $f''(x)$ ?
Answer: $f''(x)=24x(2x^3-1)+72x^4$ Chain rule on squares yields higher derivatives via product rule. The first is f'(x) = $12x^2$ ( $2x^3$ - 1). The second is f''(x) = 24x ( $2x^3$ - 1) + $72x^4$ , expanding properly. A common stopping error is not applying product rule for the second. Simplify by factoring if possible. A transferable strategy is to treat as composition and differentiate step-by-step, verifying with expansion.
If $f(x)=\frac{1}{3}x^3-4x$ , what is the third derivative $f^{(3)}(x)$ ?
Answer: $f^{(3)}(x)=2$ Higher-order derivatives involve differentiating a function multiple times, here up to the third derivative for the polynomial ( $f(x) = \frac{1}{3}x^3 - 4x$ ). Start by finding the first derivative: ( f'(x) = $x^2$ - 4 ), then the second: ( f''(x) = 2x ), and finally the third: ( f'''(x) = 2 ), which matches choice A. Each differentiation reduces the degree of the polynomial by one, leading to a constant for the third derivative. A common error is stopping at the second derivative and mistaking ( 2x ) for the third, but successive differentiation must continue until the required order. Another pitfall is misapplying the power rule, like forgetting to divide by the exponent properly. To master higher-order derivatives, always verify each step by differentiating term by term and checking the order required.
If $r(x)=\ln(x)+x^2$ , what is the second derivative $r''(x)$ for $x>0$ ?
Answer: $r''(x)=2-\frac{1}{x^2}$ Higher-order derivatives of logarithmic and polynomial terms require combining rules. The first is r'(x) = 1/x + 2x. The second is r''(x) = -1/ $x^2$ + 2, with the logarithmic part contributing the negative. A common error is stopping after the first or forgetting the domain x > 0. Differentiate each part separately. For such functions, differentiate term by term repeatedly, ensuring domain restrictions are noted.

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