Derivatives of Reciprocal Trig Functions - AP Calculus AB

A model defines $g(x)=\csc(x)\cot(x)$ . What is $g'(x)$ ?
Answer: $-\csc(x)\cot^2(x)-\csc^3(x)$ This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. g(x) = csc(x) cot(x), product rule: csc' cot + csc cot'. csc' = -csc cot, cot' = -csc², so -csc cot cot + csc (-csc²) = -csc cot² - csc³. Choice B omits negatives, tempting without signs, but both derivatives contribute negatives. Remember negatives in csc and cot derivatives compound in products.
For $f(x)=\csc(2x+5)$ , what is the derivative $f'(x)$ ?
Answer: $-2\csc(2x+5)\cot(2x+5)$ This problem tests the differentiation of cosecant, one of the reciprocal trigonometric functions. The derivative of csc(u) is -csc(u)cot(u)·u', where the negative sign is crucial and u' represents the derivative of the inner function. For f(x) = csc(2x+5), we have u = 2x+5, giving u' = 2. Thus f'(x) = -csc(2x+5)cot(2x+5)·2 = -2csc(2x+5)cot(2x+5). Students often forget the negative sign, incorrectly writing 2csc(2x+5)cot(2x+5). When differentiating csc and cot, always remember they produce negative derivatives, unlike sec and tan which give positive results.
For the temperature adjustment $T(x)=\csc(x^2)$ , what is $T'(x)$ ?
Answer: $-2x\csc(x^2)\cot(x^2)$ This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cosecant function. The function is T(x) = csc(x²), so let u = x² with u' = 2x. The derivative of csc(u) is -csc(u) cot(u) multiplied by u', yielding -2x csc(x²) cot(x²). This demonstrates the negative sign inherent in the cosecant derivative combined with the chain rule. A tempting distractor is choice C, which forgets the 2x factor from the chain rule, ignoring the inner quadratic. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.
Let $S(x)=\sec(x^2-x)$ . What is $S'(x)$ ?
Answer: $(2x-1)\sec(x^2-x)\tan(x^2-x)$ This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The function is S(x) = sec(x² - x), so let u = x² - x with u' = 2x - 1. The derivative of sec(u) is sec(u) tan(u) multiplied by u', resulting in (2x - 1) sec(x² - x) tan(x² - x). This uses the quadratic inner derivative. A tempting distractor is choice B, which forgets the 2x - 1 chain factor. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.
Let $W(x)=\sec(x)/2$ . What is $W'(x)$ ?
Answer: $\dfrac{1}{2}\sec(x)\tan(x)$ This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. W(x) = sec(x)/2, so derivative is (1/2) * sec(x) tan(x), since constant multiples factor out. No chain rule here as argument is simple. Thus, W'(x) = (1/2) sec(x) tan(x). Choice A forgets the 1/2, tempting if ignoring the constant, but constants scale the derivative. Remember to handle coefficients and constants before applying trig derivative rules.
A function is $E(x)=\csc(x^2-x+4)$ . What is $E'(x)$ ?
Answer: $-(2x-1)\csc(x^2-x+4)\cot(x^2-x+4)$ This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. For E(x) = csc(x² - x + 4), the derivative is -csc(u) cot(u) u' where u = x² - x + 4. Then u' = 2x - 1. So E'(x) = -csc(x² - x + 4) cot(x² - x + 4) * (2x - 1). Choice E substitutes csc² incorrectly, tempting if mixing with cot's derivative, but csc uses csc cot with a negative. Consistently recall that csc and cot derivatives both have negatives, but their forms differ.
If $p(x)=\sec(x^2)$ , what is $p'(x)$ ?
Answer: $2x\sec(x^2)\tan(x^2)$ This question tests differentiating secant with a non-linear argument, requiring both knowledge of reciprocal trig derivatives and the chain rule. The derivative of sec(u) is sec(u)tan(u)·u', where u' is found using the chain rule. For p(x) = sec(x²), we have u = x², so u' = 2x. Therefore, p'(x) = sec(x²)tan(x²)·2x = 2x·sec(x²)tan(x²). Students might incorrectly write just sec(x²)tan(x²), forgetting to multiply by the derivative of x². Always remember to apply the chain rule when the argument of a trig function is anything other than a simple x.
A temperature adjustment is modeled by $h(t)=\cot(5t)$ . What is $h'(t)$ ?
Answer: $-5\csc^2(5t)$ This question tests knowledge of the cotangent derivative, another reciprocal trigonometric function. The derivative of cot(u) is -csc²(u) times the derivative of u. Since u = 5t, applying the chain rule gives h'(t) = -csc²(5t) · 5 = -5csc²(5t). Students might confuse this with the tangent derivative and write 5sec²(5t) (choice D), forgetting that cotangent pairs with cosecant squared, not secant squared. The key pattern is that tan' = sec² while cot' = -csc², with cotangent carrying a negative sign.
For the function $a(t)=\sec(t^3)$ , what is $a'(t)$ ?
Answer: $3t^2\sec(t^3)\tan(t^3)$ This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The function is a(t) = sec(t³), so let u = t³ with u' = 3t². The derivative of sec(u) is sec(u) tan(u) multiplied by u', giving 3t² sec(t³) tan(t³). This applies chain rule to a cubic inner. A tempting distractor is choice B, which omits the 3t² factor. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.
A function is $B(x)=\cot(e^x)$ . What is $B'(x)$ ?
Answer: $-e^x\csc^2(e^x)$ This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. For B(x) = cot( $e^x$ ), use -csc²(u) u' with u = $e^x$ . u' = $e^x$ . So B'(x) = -csc²( $e^x$ ) * $e^x$ = - $e^x$ csc²( $e^x$ ). Choice E confuses with csc cot form, which is for csc, not cot, but cot uses csc². Distinguish between the derivative patterns: csc² for cot, sec tan for sec, and csc cot for csc.
Let $g(x)=\cot(4x)$ . Find $g'(x)$ .
Answer: $-4\csc^2(4x)$ This question involves differentiating the cotangent function, another reciprocal trigonometric function. The derivative of cot(u) is -csc²(u)·u', where the negative sign and squared cosecant are both essential. For g(x) = cot(4x), we identify u = 4x, so u' = 4. Therefore, g'(x) = -csc²(4x)·4 = -4csc²(4x). A tempting error is confusing this with the derivative of tangent and writing 4sec²(4x), but cotangent's derivative involves cosecant squared, not secant squared. Remember the pattern: tan' = sec², cot' = -csc², and both cotangent and cosecant derivatives are negative.
A model uses $m(x)=\sec(2-3x)$ . What is $m'(x)$ ?
Answer: $-3\sec(2-3x)\tan(2-3x)$ This problem tests differentiating secant with a linear decreasing argument, requiring careful application of the chain rule. The derivative of sec(u) is sec(u)tan(u)·u', and for m(x) = sec(2-3x), we have u = 2-3x, giving u' = -3. Thus m'(x) = sec(2-3x)tan(2-3x)·(-3) = -3sec(2-3x)tan(2-3x). A tempting mistake would be writing 3sec(2-3x)tan(2-3x), missing the negative sign from differentiating (2-3x). Remember that when the inner function has a negative coefficient on x, this introduces a negative sign in the final derivative.
A cost adjustment is $k(x)=x\sec(x)$ . What is $k'(x)$ ?
Answer: $\sec(x)+x\sec(x)\tan(x)$ This problem requires the product rule to differentiate x·sec(x). Using the product rule: k'(x) = [x]'sec(x) + x[sec(x)]' = 1·sec(x) + x·sec(x)tan(x) = sec(x) + x·sec(x)tan(x). A common mistake would be to forget the first term from differentiating x, yielding just x·sec(x)tan(x) (choice C), or to misapply the secant derivative formula. When using the product rule with reciprocal trig functions, remember to include both terms and apply the correct derivative formula for each reciprocal function.
Let $A(x)=\csc(\ln x)$ . What is $A'(x)$ ?
Answer: $-\dfrac{1}{x}\csc(\ln x)\cot(\ln x)$ This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. A(x) = csc(ln x), so derivative is -csc(u) cot(u) u' where u = ln x. u' = 1/x. Therefore, A'(x) = -csc(ln x) cot(ln x) * (1/x) = - (1/x) csc(ln x) cot(ln x). Choice C forgets the 1/x factor, a common chain rule omission, but logarithmic inner functions require it. Practice identifying the inner function and its derivative to master composite trig derivatives.
In a geometry model, $h(\theta)=\sec(\theta-4)$ . What is $h'(\theta)$ ?
Answer: $\sec(\theta-4)\tan(\theta-4)$ This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The function is $h(\theta) = \sec(\theta - 4)$ , so let $u = \theta - 4$ with $u' = 1$ . The derivative of $\sec(u)$ is $\sec(u) \tan(u)$ multiplied by $u'$ , simplifying to $\sec(\theta - 4) \tan(\theta - 4)$ . This structure shows the basic secant derivative with a simple chain. A tempting distractor is choice B, which adds an unnecessary negative sign, perhaps confusing with cosecant. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.
Let $D(x)=\sec(\sin x)$ . What is $D'(x)$ ?
Answer: $\cos x\,\sec(\sin x)\tan(\sin x)$ This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. D(x) = sec(sin x), derivative is sec(u) tan(u) u' where u = sin x. u' = cos x. Thus, D'(x) = sec(sin x) tan(sin x) * cos x. Choice A misses the cos x factor, tempting without chain rule, but composition requires it. When trig functions are nested, apply chain rule multiple times if necessary for accurate derivatives.
Let $y=\cot(x)$ . What is $\dfrac{dy}{dx}$ ?
Answer: $-\csc^2(x)$ This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cotangent function. The function is y = cot(x), with a simple argument. The derivative is -csc²(x), recalling the basic form. This structure is foundational without chain rule complexity. A tempting distractor is choice A, which omits the negative and confuses with tangent's derivative. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.
A model uses $F(x)=\csc(7x)$ . What is $F'(x)$ ?
Answer: $-7\csc(7x)\cot(7x)$ This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cosecant function. The function is F(x) = csc(7x), so let u = 7x with u' = 7. The derivative of csc(u) is -csc(u) cot(u) multiplied by u', resulting in -7 csc(7x) cot(7x). This scales by the constant 7. A tempting distractor is choice C, which forgets the 7 from the chain rule. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.
A signal is modeled by $p(t)=\csc(t^2+4)$ . What is $p'(t)$ ?
Answer: $-2t\csc(t^2+4)\cot(t^2+4)$ This question involves differentiating cosecant with a polynomial argument. The derivative of csc(u) is -csc(u)cot(u) times the derivative of u. For p(t) = csc(t²+4), we apply the chain rule: p'(t) = -csc(t²+4)cot(t²+4) · 2t = -2t·csc(t²+4)cot(t²+4). A common error is dropping the negative sign (choice B), forgetting that cosecant's derivative always includes a negative. Remember that the reciprocal trig derivatives follow patterns: sec pairs with tan (both positive), while csc pairs with cot (both negative).
In a model, $P(t)=\sec(t-\pi/6)+1$ . What is $P'(t)$ ?
Answer: $\sec(t-\pi/6)\tan(t-\pi/6)$ This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the secant function. The function is P(t) = sec(t - π/6) + 1, so the constant 1 differentiates to 0. Let u = t - π/6 with u' = 1, so the derivative is sec(u) tan(u) multiplied by 1, or sec(t - π/6) tan(t - π/6). This ignores the added constant. A tempting distractor is choice C, which confuses with tangent's derivative. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

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