Differentiating Inverse Trigonometric Functions - AP Calculus BC

In a model, $g(x)=\arctan(5x-1)$ . What is $g'(x)$ ?
Answer: $\dfrac{5}{1+(5x-1)^2}$ This problem involves differentiating the inverse tangent function, a key skill in AP Calculus BC. The derivative formula for arctan(u) is u' divided by 1 plus u squared. Here, u equals 5x minus 1, so u' is 5. Applying the formula gives 5 over 1 plus (5x minus 1) squared. A tempting distractor is choice A, which uses the arcsin formula with a square root instead of the correct arctan denominator. Always remember to apply the chain rule correctly and verify the sign based on the specific inverse trig function's derivative formula.
For a signal phase shift, $p(x)=\operatorname{arcsec}(2x+1)$ . What is $p'(x)$ ?
Answer: $\dfrac{2}{|2x+1|\sqrt{(2x+1)^2-1}}$ This problem involves differentiating arcsec, one of the less common inverse trig functions. The derivative of arcsec(u) is u'/(|u|√(u²-1)), where u = 2x + 1. Using the chain rule, u' = 2. Therefore, p'(x) = 2/(|2x+1|√((2x+1)²-1)), which is choice C. Choice A incorrectly uses the arcsin formula, while choice E omits the absolute value. For arcsec and arccsc, always include |u| in the denominator along with √(u²-1).
A sonar angle is $y(x)=\arctan\!\left(\frac{1}{x}\right)$ for $x\ne 0$ . What is $y'(x)$ ?
Answer: $\dfrac{-1/x^2}{1+\left(\frac{1}{x}\right)^2}$ This problem involves differentiating arctangent with a reciprocal argument. The derivative of arctan(u) is 1/(1+u²), and we apply the chain rule with u = 1/x. Differentiating: y'(x) = 1/(1+(1/x)²) · d/dx(1/x) = 1/(1+1/x²) · (-1/x²) = -1/x²/(1+1/x²). Choice C incorrectly uses the arcsin formula with √(1-(1/x)²) in the denominator. For inverse trig derivatives, remember that arctangent uses 1+u² without any square root, distinguishing it from arcsin and arccos formulas.
A signal phase shift is $q(x)=\arccsc(x^3)$ . What is $q'(x)$ for $x\neq 0$ ?
Answer: $\dfrac{-3x^2}{|x^3|\sqrt{x^6-1}}$ Differentiating inverse trigonometric functions is the key skill here, specifically finding the derivative of arccsc( $x^3$ ). The formula for the derivative of arccsc(u) where u is a function of x is (-1 / (|u| sqrt( $u^2$ - 1))) times u'. In this case, u = $x^3$ , so u' = $3x^2$ , leading to q'(x) = - $3x^2$ / (| $x^3$ | sqrt( $(x^3)^2$ - 1)). Simplifying, that's - $3x^2$ / (| $x^3$ | sqrt( $x^6$ - 1)), which matches choice A. A tempting distractor like choice B fails because it omits the negative sign essential to the arccsc derivative. Always remember to apply the chain rule when differentiating composite inverse trig functions and double-check the sign based on the specific function.
In a modeling lab, $v(x)=\arccsc(2x-3)$ . What is $v'(x)$ ?
Answer: $\dfrac{-2}{|2x-3|\sqrt{(2x-3)^2-1}}$ Differentiating inverse trigonometric functions is the key skill here, specifically finding the derivative of arccsc(2x-3). The formula for the derivative of arccsc(u) where u is a function of x is (-1 / (|u| sqrt( $u^2$ - 1))) times u'. In this case, u = 2x-3, so u' = 2, leading to v'(x) = -2 / (|2x-3| sqrt( $(2x-3)^2$ - 1)). This matches choice C directly. A tempting distractor like choice B fails because it omits the negative sign required for arccsc. Always remember to apply the chain rule when differentiating composite inverse trig functions and double-check the sign based on the specific function.
A control function is $h(x)=\arccos(5x-2)$ . What is $h'(x)$ ?
Answer: $-\dfrac{5}{\sqrt{1-(5x-2)^2}}$ This problem requires differentiating arccos with a linear argument. The derivative of arccos(u) is -u'/√(1-u²), where u = 5x - 2. The chain rule gives u' = 5. Therefore, h'(x) = -5/√(1-(5x-2)²), which matches choice B. Choice A omits the negative sign that is essential for arccos derivatives. Choice C incorrectly uses the arctan formula. Remember that arccos always has a negative sign in its derivative, distinguishing it from arcsin.
For a timing offset, $s(x)=\arcsin(4x^2)$ . What is $s'(x)$ ?
Answer: $\dfrac{8x}{\sqrt{1-(4x^2)^2}}$ This problem requires differentiating arcsin with a quadratic argument. The derivative of arcsin(u) is u'/√(1-u²), where u = 4x². Using the chain rule, u' = 8x. Therefore, s'(x) = 8x/√(1-(4x²)²) = 8x/√(1-16x⁴), which is choice A. Choice B incorrectly uses the arctan formula, while choice C includes an erroneous negative sign. When differentiating arcsin, ensure you use √(1-u²) in the denominator with no negative sign.
A glide path is modeled by $w(x)=\arccos(x^2-1)$ . What is $w'(x)$ ?
Answer: $-\dfrac{2x}{\sqrt{1-(x^2-1)^2}}$ This problem requires differentiating arccos with a quadratic composite function. The derivative of arccos(u) is -1/√(1-u²), and we need the chain rule with u = x²-1. Differentiating: w'(x) = -1/√(1-(x²-1)²) · d/dx(x²-1) = -1/√(1-(x²-1)²) · 2x = -2x/√(1-(x²-1)²). Choice A incorrectly omits the negative sign from the arccos derivative, giving 2x/√(1-(x²-1)²). When differentiating inverse trig functions, arccos uniquely includes a negative sign that must not be forgotten.
In a model, $g(x)=\arcsin\!\left(\dfrac{2x}{1+x^2}\right)$ . What is $g'(x)$ ?
Answer: $\dfrac{\frac{2(1-x^2)}{(1+x^2)^2}}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}$ This problem involves differentiating arcsin with a rational function argument. The derivative of arcsin(u) is $\frac{u'}{\sqrt{1-u^2}}$ , where $u = \frac{2x}{1+x^2}$ . Using the quotient rule, $u' = \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2}$ . Substituting into the arcsin derivative formula gives $g'(x) = \frac{\frac{2(1-x^2)}{(1+x^2)^2}}{\sqrt{1 - \left( \frac{2x}{1+x^2} \right)^2}}$ , which is choice A. Choice B incorrectly uses the arctan formula with $1 + u^2$ instead of $\sqrt{1 - u^2}$ . For arcsin derivatives, always use $\sqrt{1 - u^2}$ in the denominator.
A calibration curve is $t(x)=\arccos(x^2)$ . What is $t'(x)$ for $|x|<1$ ?
Answer: $\dfrac{-2x}{\sqrt{1-x^4}}$ Differentiating inverse trigonometric functions is the key skill here, specifically finding the derivative of arccos( $x^2$ ). The formula for the derivative of arccos(u) where u is a function of x is (-1 / sqrt(1 - $u^2$ )) times u'. In this case, u = $x^2$ , so u' = 2x, leading to t'(x) = -2x / sqrt(1 - $(x^2)^2$ ). Simplifying, that's -2x / sqrt(1 - $x^4$ ), which matches choice B. A tempting distractor like choice A fails because it omits the negative sign specific to arccos. Always remember to apply the chain rule when differentiating composite inverse trig functions and double-check the sign based on the specific function.
A control signal is $q(x)=\arcsin\!\left(\frac{x}{2}\right)$ . What is $q'(x)$ ?
Answer: $\dfrac{1}{2\sqrt{1-\left(\frac{x}{2}\right)^2}}$ This problem requires differentiating arcsin with a fractional linear argument. The derivative of arcsin(u) is 1/√(1-u²), and we need the chain rule with u = x/2. Differentiating: q'(x) = 1/√(1-(x/2)²) · d/dx(x/2) = 1/√(1-x²/4) · (1/2) = 1/(2√(1-x²/4)). Choice D incorrectly applies the arctangent formula without the square root, giving 1/(1+(x/2)²). When differentiating inverse trig functions with fractional arguments, carefully track how the chain rule affects both the denominator and the overall coefficient.
For $g(x)=\arcsin(5x-1)$ , what is $g'(x)$ ?
Answer: $\dfrac{5}{\sqrt{1-(5x-1)^2}}$ This problem involves differentiating arcsine of a linear expression. The derivative of arcsin(u) is 1/√(1-u²) times the derivative of u. With u = 5x-1, we have u' = 5. Using the chain rule: g'(x) = 1/√(1-(5x-1)²) · 5 = 5/√(1-(5x-1)²). Choice C incorrectly includes a negative sign, which would apply to arccos but not arcsin. Remember that arcsin has a positive derivative formula, while arccos includes a negative sign.
A model uses $h(x)=\arccos(2x+3)$ . What is $h'(x)$ ?
Answer: $-\dfrac{2}{\sqrt{1-(2x+3)^2}}$ This requires differentiating arccos, which has the derivative formula -1/√(1-u²) times u'. Here, u = 2x+3, so u' = 2. Applying the chain rule: h'(x) = -1/√(1-(2x+3)²) · 2 = -2/√(1-(2x+3)²). Choice A omits the negative sign that is essential to the arccos derivative formula. The negative sign in arccos derivatives reflects that cosine decreases as its angle increases in the range [0,π]. Always include the negative sign when differentiating arccos.
A lens model defines $t(x)=\arccos\!\left(\dfrac{1}{x}\right)$ for $x>1$ . What is $t'(x)$ ?
Answer: $-\dfrac{1/x^2}{\sqrt{1-(1/x)^2}}$ This problem involves differentiating arccos with a reciprocal argument. The derivative of arccos(u) is -u'/√(1-u²), where u = 1/x. Using the power rule, u' = -1/x². Therefore, t'(x) = -(-1/x²)/√(1-(1/x)²) = (1/x²)/√(1-1/x²), which matches choice B. Choice A omits the negative sign from arccos, while choice C uses the arctan formula. The double negative from arccos and the reciprocal's derivative creates a positive result.
A tracking model uses $m(x)=\arctan\!\left(\sqrt{x}\right)$ . What is $m'(x)$ ?
Answer: $\dfrac{1}{2\sqrt{x}\,\left(1+\left(\sqrt{x}\right)^2\right)}$ This problem requires differentiating arctangent with a square root composite function. The derivative of arctan(u) is 1/(1+u²), and we need the chain rule with u = √x. Differentiating: m'(x) = 1/(1+(√x)²) · d/dx(√x) = 1/(1+x) · 1/(2√x) = 1/(2√x(1+x)). Choice A incorrectly uses the arcsin formula with a square root in the denominator, giving 1/(2√x·√(1+(√x)²)). When differentiating inverse trig functions with radical arguments, remember that arctangent derivatives never contain square roots except those from the chain rule.
A signal is $u(x)=\arccot(1-x)$ . What is $u'(x)$ ?
Answer: $\dfrac{1}{1+(1-x)^2}$ This requires differentiating arccot, which has the formula -1/(1+u²) times u'. With u = 1-x, we have u' = -1. Applying the chain rule: u'(x) = -1/(1+(1-x)²) · (-1) = 1/(1+(1-x)²). Choice B incorrectly uses the arcsin formula with a square root, but arccot derivatives have no radical. The double negative (from the arccot formula and the derivative of 1-x) results in a positive answer. For arccot, remember the formula is -1/(1+u²), not involving any square root.
A signal phase is $n(t)=\arcsin\!\left(\frac{2}{t}\right)$ for $t>2$ . What is $n'(t)$ ?
Answer: $\dfrac{-2/t^2}{\sqrt{1-\left(\frac{2}{t}\right)^2}}$ This problem involves differentiating arcsin with a reciprocal function. The derivative of arcsin(u) is 1/√(1-u²), and we apply the chain rule with u = 2/t. Differentiating: n'(t) = 1/√(1-(2/t)²) · d/dt(2/t) = 1/√(1-4/t²) · (-2/t²) = -2/t²/√(1-4/t²). Choice B incorrectly has a positive result, missing the negative sign from differentiating 2/t. For inverse trig derivatives with reciprocal arguments, remember that d/dx(1/x) = -1/x², introducing a negative sign.
A sensor model uses $f(x)=\arcsin(3x)$ for $|x|<\tfrac13$ . What is $f'(x)$ ?
Answer: $\dfrac{3}{\sqrt{1-9x^2}}$ Differentiating inverse trigonometric functions is the key skill here, specifically finding the derivative of arcsin(3x). The formula for the derivative of arcsin(u) where u is a function of x is (1 / sqrt(1 - $u^2$ )) times u'. In this case, u = 3x, so u' = 3, leading to f'(x) = 3 / sqrt(1 - $(3x)^2$ ). Simplifying, that's 3 / sqrt(1 - $9x^2$ ), which matches choice A. A tempting distractor like choice C fails because it includes a negative sign, which would apply to arccos but not arcsin. Always remember to apply the chain rule when differentiating composite inverse trig functions and double-check the sign based on the specific function.
In an optics experiment, $q(x)=\arccsc(x^3)$ . What is $q'(x)$ ?
Answer: $\dfrac{-3x^2}{|x^3|\sqrt{x^6-1}}$ This question requires differentiating the inverse cosecant function with a power argument. The derivative of arccsc(u) is -1/(|u|√(u²-1)) · u', where u = x³. Applying the chain rule gives d/dx[arccsc(x³)] = -1/(|x³|√((x³)²-1)) · d/dx[x³] = -1/(|x³|√(x⁶-1)) · 3x² = -3x²/(|x³|√(x⁶-1)). Choice E incorrectly uses the positive version instead of the negative derivative for arccsc. Like arccos, arccsc has a negative derivative, distinguishing it from arcsec.
In a control system, $h(x)=\arccos(2x-3)$ . What is $h'(x)$ ?
Answer: $\dfrac{-2}{\sqrt{1-(2x-3)^2}}$ This question requires differentiating the inverse cosine function with a composite argument. The derivative of arccos(u) is -1/√(1-u²) · u', where u = 2x-3. Applying the chain rule gives d/dx[arccos(2x-3)] = -1/√(1-(2x-3)²) · d/dx[2x-3] = -1/√(1-(2x-3)²) · 2 = -2/√(1-(2x-3)²). Choice A incorrectly uses the positive version (like arcsin) instead of the negative derivative for arccos. Remember that arccos has a negative derivative while arcsin has a positive one.

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