Derivatives of Trigonometry and Logarithmic Functions - AP Calculus AB

A function is $w(x)=\ln\!\left(\sin\!\left(x^2+1\right)\right)$ . What is $w'(x)$ ?
Answer: $\dfrac{2x\cos(x^2+1)}{\sin(x^2+1)}$ To find w'(x) for w(x) = ln(sin(x² + 1)), this is log of trig, so chain rule. Outer ln(b), derivative 1/b, b = sin(x² + 1). b' = cos(x² + 1) 2x. So w'(x) = (1/sin(x² + 1)) cos(x² + 1) * 2x = 2x cot(x² + 1). A common omission is forgetting the 2x from the inner quadratic's derivative. Pattern strategy: Identify logs of trig with inner polynomials and apply chain rule to both the trig and its argument.
For $m(x)=\ln\!\left(\sqrt{1+\sin x}\right)$ , what is $m'(x)$ ?
Answer: $\dfrac{\cos x}{2(1+\sin x)}$ We can simplify first: ln(√(1+sin x)) = ln((1+sin x)^(1/2)) = (1/2)ln(1+sin x). Now differentiating: m'(x) = (1/2) · 1/(1+sin x) · cos x = cos x/(2(1+sin x)). Alternatively, treating it as a composition: derivative of ln(u) is 1/u, so 1/√(1+sin x), times the derivative of √(1+sin x), which is cos x/(2√(1+sin x)). This gives cos x/(2(1+sin x)) after simplification. Students often forget to apply logarithm properties or make errors with the square root derivative. Pattern tip: ln(√(expression)) can be simplified using log properties before differentiating, often making the work easier.
Let $m(x)=\log_{10}\!\left(2+\sin x\right)$ . What is $m'(x)$ ?
Answer: $\dfrac{1}{\ln 10}\cdot\dfrac{\cos x}{2+\sin x}$ For m(x) = log₁₀(2 + sin x), we need the chain rule combined with the base-10 logarithm derivative formula. The derivative of log₁₀(u) is 1/(u·ln 10), so we get 1/((2 + sin x)·ln 10) for the outer derivative. The inner function 2 + sin x has derivative 0 + cos x = cos x. Multiplying outer and inner derivatives: m'(x) = [1/((2 + sin x)·ln 10)]·cos x = cos x/((2 + sin x)·ln 10) = (1/ln 10)·[cos x/(2 + sin x)]. A common mistake is treating log₁₀ like ln and forgetting the factor of 1/ln 10. When differentiating logarithms with bases other than e, always include the conversion factor 1/ln(base) in your derivative.
For $s(x)=\sin\!\left((3x-1)^4\right)$ , what is $s'(x)$ ?
Answer: $12(3x-1)^3\cos\!\left((3x-1)^4\right)$ To differentiate s(x) = sin((3x - 1)⁴), we have a three-layer composition: sine on the outside, fourth power in the middle, and linear expression innermost. The outer derivative of sin(u) is cos(u), giving cos((3x - 1)⁴). The middle layer u⁴ has derivative 4u³, contributing 4(3x - 1)³. The innermost layer (3x - 1) differentiates to 3. Multiplying all three: s'(x) = cos((3x - 1)⁴)·4(3x - 1)³·3 = 12(3x - 1)³cos((3x - 1)⁴). Students often miss one of the layers, especially forgetting the factor of 3 from the innermost derivative. For trig(power(linear)), work methodically through each layer: trig → power → linear, multiplying all derivative factors.
A function is $B(x)=\tan\!\left(\ln\!\left(\dfrac{1}{x}\right)\right)$ . What is $B'(x)$ ?
Answer: $-\dfrac{1}{x}\sec^2\!\left(\ln\!\left(\dfrac{1}{x}\right)\right)$ To differentiate B(x) = tan(ln(1/x)), apply chain rule. Outer tan(e), derivative sec²(e), e = ln(1/x) = -ln x. e' = -1/x. So B'(x) = sec²(e) * (-1/x). A common omission is forgetting the negative from the inner log's derivative. Pattern strategy: Note trig of log of reciprocal and simplify log first to capture the sign in the chain.
Let $N(x)=\tan\!\left(\dfrac{2}{x^2-1}\right)$ . What is $N'(x)$ ?
Answer: $-\dfrac{4x}{(x^2-1)^2}\sec^2\!\left(\dfrac{2}{x^2-1}\right)$ For N(x) = tan(2/(x² - 1)), chain rule. Outer tan(r), derivative sec²(r), r = 2 $(x² - 1)^{-1}$ . r' = 2 * (-1) $(x² - 1)^{-2}$ 2x = -4x / (x² - 1)². Thus, N'(x) = sec²(r) (-4x / (x² - 1)²). A common omission is coefficient error in r'. Recognize trig of rational with quadratic denominator and use power chain.
A function is defined by $s(x)=\ln\!\left((x^2+1)^3\right)$ . What is $s'(x)$ ?
Answer: $\dfrac{6x}{x^2+1}$ This log of a power function simplifies via log properties but requires chain rule. s(x) = 3 ln( $x^2$ + 1), so s'(x) = 3 * (2x)/( $x^2$ + 1) = 6x/( $x^2$ + 1). Outer ln, inner $(x^2 + 1)^3$ , but property avoids direct chain on power. Common omission: not multiplying by 3 after simplifying. Forgetting 2x in numerator is frequent. Recognize log of power as multiple of log and differentiate accordingly. This pattern transfers to exponential logs, enhancing efficiency.
A function is $j(x)=\ln\!\left(\dfrac{1}{x^2+2}\right)$ . What is $j'(x)$ ?
Answer: $\dfrac{-2x}{x^2+2}$ Log of reciprocal polynomial uses properties. j(x) = -ln( $x^2$ + 2), j'(x) = - (2x)/( $x^2$ + 2). Outer ln, inner 1/( $x^2$ + 2), but property simplifies. Common omission: sign error from reciprocal. Missing 2x. Identify log of inverse and use negative log then differentiate. This pattern aids in denominator logs.
A rate is given by $u(x)=\tan\!\left(\ln(x^2+4)\right)$ . What is $u'(x)$ ?
Answer: $\dfrac{2x}{x^2+4}\sec^2\!\left(\ln(x^2+4)\right)$ This tangent of a log requires chain rule for trig-log composite. Outer tan(u), u = ln( $x^2$ + 4), derivative $\sec^2$ (u), inner u' = 2x/( $x^2$ + 4). So u'(x) = (2x/( $x^2$ + 4)) $\sec^2$ (ln( $x^2$ + 4)). Common omission: forgetting 2x from inner derivative. Confusing tan derivative with sec tan is another error. Recognize trig of log polynomial and chain fully. This pattern aids in complex nests, building derivative skills.
Let $L(x)=\cos\!\left(\ln\!\left(\sqrt{1+x}\right)\right)$ . What is $L'(x)$ ?
Answer: $-\dfrac{\sin\!\left(\ln(\sqrt{1+x})\right)}{2(1+x)}$ For L(x) = cos(ln(√(1 + x))), simplify inner: ln(√(1 + x)) = (1/2) ln(1 + x). Outer cos(p), derivative -sin(p), p' = (1/2) * 1/(1 + x). Thus, L'(x) = -sin(p) / (2 (1 + x)). A common omission is missing the 1/2. Recognize log of root as scaled log for simpler chaining.
Let $g(t)=\ln\!\left(\cos(2t-1)\right)$ . What is $g'(t)$ ?
Answer: $\dfrac{-2\sin(2t-1)}{\cos(2t-1)}$ This composition involves ln as the outer function and cos(2t-1) as the inner function. The derivative of ln(u) is 1/u, so we get 1/cos(2t-1) times the derivative of the inner function. The derivative of cos(2t-1) is -sin(2t-1) times 2 (from the chain rule on 2t-1). This gives us (1/cos(2t-1)) · (-2sin(2t-1)) = -2sin(2t-1)/cos(2t-1). Students often forget the factor of 2 from differentiating the linear expression 2t-1. Remember: when differentiating ln(trig function), the result often simplifies to a trig ratio times any constants from the innermost derivative.
Let $r(x)=\sin\!\left(\dfrac{1}{x^2+3}\right)$ . What is $r'(x)$ ?
Answer: $-\dfrac{2x\cos\!\left(\dfrac{1}{x^2+3}\right)}{(x^2+3)^2}$ Differentiating sine with a rational inner function uses chain rule. Outer sin(u), u = 1/( $x^2$ + 3), derivative cos(u), inner u' = -2x/ $(x^2 + 3)^2$ . So r'(x) = cos(1/( $x^2$ + 3)) * (-2x/ $(x^2 + 3)^2$ ). Common omission: forgetting the negative from the reciprocal's derivative. Also, missing the squared denominator in u'. Spot trig of reciprocal polynomial and chain step-by-step. This strategy applies to rational composites, aiding verification.
Let $A(x)=\sin\!\left(2\ln(3x)\right)$ . What is $A'(x)$ ?
Answer: $\dfrac{2}{x}\cos\!\left(2\ln(3x)\right)$ For A(x) = sin(2 ln(3x)), chain rule for trig of log. Outer sin(d), derivative cos(d), d = 2 ln(3x). d' = 2 (1/(3x)) 3 = 2/x. Thus, A'(x) = cos(d) * (2/x). A common omission is miscalculating the log derivative, forgetting the chain for ln(3x). Recognize patterns in trig of scaled logs by differentiating the scalar multiple and log chain.
If $h(x)=\tan\!\left(\sqrt{5x+1}\right)$ , what is $h'(x)$ ?
Answer: $\sec^2\!\left(\sqrt{5x+1}\right)\cdot\dfrac{5}{2\sqrt{5x+1}}$ Here we have tan composed with a square root function. The derivative of tan(u) is sec²(u), and we must multiply by the derivative of √(5x+1). Using the chain rule on the square root: d/dx[√(5x+1)] = 1/(2√(5x+1)) · 5 = 5/(2√(5x+1)). Therefore, h'(x) = sec²(√(5x+1)) · 5/(2√(5x+1)). A common mistake is forgetting the factor of 5 from the derivative of 5x+1 inside the square root. Pattern recognition tip: when differentiating tan(√(expression)), expect sec² of the same argument multiplied by a fraction containing the square root in the denominator.
A function is $K(x)=\sin\!\left(\dfrac{1}{1+\cos x}\right)$ . What is $K'(x)$ ?
Answer: $\dfrac{\sin x}{(1+\cos x)^2}\cos\!\left(\dfrac{1}{1+\cos x}\right)$ To differentiate $K(x) = \sin\left(\dfrac{1}{1 + \cos x}\right)$ , chain rule. Outer $\sin(n)$ , derivative $\cos(n)$ , $n = (1 + \cos x)^{-1}$ . $n' = - (1 + \cos x)^{-2} \cdot (-\sin x) = \dfrac{\sin x}{(1 + \cos x)^2}$ . So $K'(x) = \cos(n) \cdot \dfrac{\sin x}{(1 + \cos x)^2}$ . A common omission is sign error in $n'$ . Pattern: Trig of reciprocal trig needs careful power chain.
If $c(x)=\ln\!\left(\cos(2x)\right)$ , what is $c'(x)$ ?
Answer: $-2\tan(2x)$ Log of cosine with linear argument uses chain. Outer ln(u), u = cos(2x), derivative 1/u, inner u' = -sin(2x)*2 = -2 sin(2x). c'(x) = -2 sin(2x)/cos(2x) = -2 tan(2x). Common omission: forgetting the 2 from inner. Missing negative sign. Identify log of trig and chain with care. This pattern aids in tan equivalents.
A function is $s(x)=\cos\!\left(\sqrt{2x^2-3x}\right)$ . What is $s'(x)$ ?
Answer: $-\dfrac{(4x-3)\sin\!\left(\sqrt{2x^2-3x}\right)}{2\sqrt{2x^2-3x}}$ To differentiate s(x) = cos(√(2x² - 3x)), note the multiple layers requiring repeated chain rule application. The outermost is cos(v), derivative -sin(v), where v = √(2x² - 3x). Then v's derivative is (1/(2√(2x² - 3x))) (4x - 3). Combining yields -sin(v) (4x - 3)/(2√(2x² - 3x)). A common omission is forgetting the 1/2 in the square root derivative, leading to an extra factor of 2 in the denominator. For pattern recognition, identify multi-layer compositions like trig of radical of polynomial and apply chain rule stepwise from outside in.
A signal is modeled by $p(t)=\sin\!\left(\dfrac{3}{t^2+1}\right)$ . What is $p'(t)$ ?
Answer: $\cos\!\left(\dfrac{3}{t^2+1}\right)\cdot\dfrac{-6t}{(t^2+1)^2}$ This function involves sin composed with a rational expression 3/(t²+1). The derivative of sin(u) is cos(u), so we need cos(3/(t²+1)) times the derivative of 3/(t²+1). Using the quotient rule or rewriting as 3(t²+1)⁻¹, we get: d/dt[3/(t²+1)] = -3(t²+1)⁻² · 2t = -6t/(t²+1)². Therefore, p'(t) = cos(3/(t²+1)) · (-6t/(t²+1)²). Students often make sign errors or forget to square the denominator when differentiating quotients. Remember: sin(quotient) becomes cos(quotient) times the derivative of the quotient, which typically has the denominator squared.
Let $v(x)=\ln\!\left(\sqrt{1-\sin x}\right)$ . What is $v'(x)$ ?
Answer: $-\dfrac{\cos x}{2(1-\sin x)}$ Differentiating log of square root uses chain rule after simplification. $v(x) = \frac{1}{2} \ln(1 - \sin x)$ , $v'(x) = \frac{1}{2} \cdot \frac{-\cos x}{1 - \sin x}$ . Outer ln, inner $\sqrt{1 - \sin x}$ , but simplify first. Common omission: missing the 1/2 factor. Forgetting negative from -sin x derivative. Spot log of sqrt trig and use properties then chain. This transfers to radical logs, improving accuracy.
For $f(x)=\sin\!\left(\ln(3x^2-5x+4)\right)$ , what is $f'(x)$ ?
Answer: $\cos\!\left(\ln(3x^2-5x+4)\right)\cdot\dfrac{6x-5}{3x^2-5x+4}$ This function has a three-layer composition: sine (outer), natural log (middle), and a polynomial (inner). Using the chain rule, we differentiate from outside to inside: the derivative of sin(u) is cos(u), times the derivative of ln(v) which is 1/v, times the derivative of 3x²-5x+4 which is 6x-5. Combining these gives cos(ln(3x²-5x+4)) · (1/(3x²-5x+4)) · (6x-5). A common error is forgetting to multiply by the derivative of the innermost function (6x-5). The pattern to remember: for sin(ln(polynomial)), you'll always have cos(ln(...)) multiplied by a fraction with the polynomial's derivative in the numerator and the polynomial itself in the denominator.

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