Derivatives of Trigonometry and Logarithmic Functions

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AP Calculus AB › Derivatives of Trigonometry and Logarithmic Functions

Questions 1 - 10
1

If $c(x)=\ln!\left(\cos(2x)\right)$, what is $c'(x)$?

$\ln!\left(\cos(2x)\right)\cdot(-2\sin(2x))$

$\dfrac{1}{\cos(2x)}$

$-\tan(2x)$

$-2\tan(2x)$

$\dfrac{-\sin(2x)}{\cos(2x)}$

Explanation

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.

2

Let $y(x)=\cos!\left(\dfrac{1}{\sqrt{x}}\right)$. What is $y'(x)$?

$-\dfrac{\sin!\left(\dfrac{1}{\sqrt{x}}\right)}{2x^{3/2}}$

$-\sin!\left(\dfrac{1}{\sqrt{x}}\right)$

$\dfrac{\cos!\left(\dfrac{1}{\sqrt{x}}\right)}{2x^{3/2}}$

$\dfrac{\sin!\left(\dfrac{1}{\sqrt{x}}\right)}{2x^{3/2}}$

$\dfrac{\sin!\left(\dfrac{1}{\sqrt{x}}\right)}{2\sqrt{x}}$

Explanation

For y(x) = cos(1/√x), chain rule handles the trig of reciprocal root. Outer cos(c), derivative -sin(c), c = $x^{-1/2}$. c' = (-1/2) $x^{-3/2}$. Thus, y'(x) = -sin(c) * (-1/2 $x^{-3/2}$) = sin(c) / (2 $x^{3/2}$). A common omission is missing the sign flip from the two negatives, leading to a negative result. To recognize, look for trig functions with power or root arguments and compute exponents carefully in derivatives.

3

A function is $K(x)=\sin!\left(\dfrac{1}{1+\cos x}\right)$. What is $K'(x)$?

$\dfrac{\sin x}{(1+\cos x)^2}\cos!\left(\dfrac{1}{1+\cos x}\right)$

$-\dfrac{\sin x}{1+\cos x}\cos!\left(\dfrac{1}{1+\cos x}\right)$

$\cos!\left(\dfrac{1}{1+\cos x}\right)\cdot \dfrac{1}{1+\cos x}$

$-\dfrac{\sin x}{(1+\cos x)^2}\cos!\left(\dfrac{1}{1+\cos x}\right)$

$\cos!\left(\dfrac{1}{1+\cos x}\right)$

Explanation

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.

4

Let $E(x)=\sin!\left(\ln!\left(1+x^2\right)\right)$. What is $E'(x)$?

$\dfrac{1}{1+x^2}\cos!\left(\ln(1+x^2)\right)$

$2x\cos!\left(\ln(1+x^2)\right)$

$\dfrac{2x}{1+x^2}\sin!\left(\ln(1+x^2)\right)$

$\cos!\left(\ln(1+x^2)\right)$

$\dfrac{2x}{1+x^2}\cos!\left(\ln(1+x^2)\right)$

Explanation

For E(x) = sin(ln(1 + x²)), chain rule for trig of log. Outer sin(h), derivative cos(h), h = ln(1 + x²). h' = (1/(1 + x²)) * 2x. Thus, E'(x) = cos(h) * (2x/(1 + x²)). A common omission is missing the 2x in the numerator. Recognize trig enclosing log of quadratic and chain both derivatives accurately.

5

A function is $m(x)=\ln!\left(\sqrt{x^4+5}\right)$. What is $m'(x)$?

$\dfrac{2x^3}{\sqrt{x^4+5}}$

$\dfrac{4x^3}{x^4+5}$

$\dfrac{1}{\sqrt{x^4+5}}$

$\ln!\left(\sqrt{x^4+5}\right)\cdot \dfrac{2x^3}{\sqrt{x^4+5}}$

$\dfrac{2x^3}{x^4+5}$

Explanation

Log of square root polynomial simplifies with chain. m(x) = (1/2) $ln(x^4$ + 5), m'(x) = (1/2) * $(4x^3$$)/(x^4$ + 5) = $2x^3$$/(x^4$ + 5). Common omission: missing 1/2 or 4. Incorrect power in numerator. Identify log of sqrt power and simplify then chain. This pattern helps in even powers.

6

A rate is given by $u(x)=\tan!\left(\ln(x^2+4)\right)$. What is $u'(x)$?

$\dfrac{2x}{x^2+4}\tan!\left(\ln(x^2+4)\right)$

$2x\sec^2!\left(\ln(x^2+4)\right)$

$\dfrac{1}{x^2+4}\sec^2!\left(\ln(x^2+4)\right)$

$\sec^2!\left(\ln(x^2+4)\right)$

$\dfrac{2x}{x^2+4}\sec^2!\left(\ln(x^2+4)\right)$

Explanation

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.

7

A function is $j(x)=\ln!\left(\dfrac{1}{x^2+2}\right)$. What is $j'(x)$?

$\ln!\left(\dfrac{1}{x^2+2}\right)\cdot \left(\dfrac{-2x}{(x^2+2)^2}\right)$

$\dfrac{-2x}{x^2+2}$

$\dfrac{1}{\frac{1}{x^2+2}}$

$\dfrac{2x}{x^2+2}$

$\dfrac{-2x}{(x^2+2)^2}$

Explanation

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.

8

A function is $d(x)=\sin!\left(\sqrt{3x+7}\right)$. What is $d'(x)$?

$\dfrac{3}{2\sqrt{3x+7}}\sin!\left(\sqrt{3x+7}\right)$

$\cos!\left(\sqrt{3x+7}\right)$

$\dfrac{3}{2\sqrt{3x+7}}\cos!\left(\sqrt{3x+7}\right)$

$\dfrac{3}{\sqrt{3x+7}}\cos!\left(\sqrt{3x+7}\right)$

$\dfrac{1}{2\sqrt{3x+7}}\cos!\left(\sqrt{3x+7}\right)$

Explanation

Sine of square root needs chain rule. Outer sin(u), u = √(3x + 7), derivative cos(u), inner u' = (1/2)(3x + $7)^{-1/2}$ * 3 = 3/(2 √(3x + 7)). d'(x) = cos(√(3x + 7)) * 3/(2 √(3x + 7)). Common omission: missing 1/2 or 3. Incorrect exponent handling. Spot trig of radical linear and layer chain. This transfers to root composites.

9

A sensor records $f(x)=\sin!\left((3x-2)^5\right)$. What is $f'(x)$?

$15(3x-2)^4\cos!\left((3x-2)^5\right)$

$\sin!\left((3x-2)^5\right)\cdot 15(3x-2)^4$

$5(3x-2)^4\cos!\left((3x-2)^5\right)$

$3\cos!\left((3x-2)^5\right)$

$\cos!\left((3x-2)^5\right)$

Explanation

This problem requires finding the derivative of a composite function involving a trigonometric sine function with a polynomial inner expression, necessitating the chain rule. The outer function is sin(u), where u = (3x - $2)^5$, and its derivative is cos(u), while the inner function u has derivative 5(3x - $2)^4$ * 3 = 15(3x - $2)^4$. To find f'(x), multiply cos((3x - $2)^5$) by 15(3x - $2)^4$, yielding the full derivative. A common omission is forgetting to apply the chain rule to the power function inside, such as missing the factor of 5 from the exponent or the 3 from the linear term. Another frequent mistake is confusing the derivative of sine with cosine without the negative sign, but here it's positive for sine. By recognizing the nested structure—trig function of a power of a linear expression—you can apply the chain rule in layers, starting from the outermost. This pattern of multiple compositions is common in calculus problems, so practice decomposing functions into outer and inner parts to build proficiency.

10

A function is $u(x)=\sin!\left(\dfrac{\ln x}{x}\right)$. What is $u'(x)$?

$\cos!\left(\dfrac{\ln x}{x}\right)$

$\left(\dfrac{\ln x}{x}\right)\cos!\left(\dfrac{\ln x}{x}\right)$

$\left(\dfrac{1}{x}\right)\cos!\left(\dfrac{\ln x}{x}\right)$

$\left(\dfrac{1-\ln x}{x^2}\right)\cos!\left(\dfrac{\ln x}{x}\right)$

$\left(\dfrac{\ln x-1}{x^2}\right)\cos!\left(\dfrac{\ln x}{x}\right)$

Explanation

Differentiating u(x) = sin((ln x)/x) involves chain rule for trig of a quotient. Outer is sin(z), derivative cos(z), where z = (ln x)/x. z' requires quotient rule: ( (1/x)*x - ln x * 1 ) / x² = (1 - ln x)/x². So u'(x) = cos(z) * (1 - ln x)/x². A common omission is mishandling the quotient derivative, like forgetting the negative term for ln x. For patterns, spot trig functions with logarithmic or rational arguments and compute inner derivatives carefully using appropriate rules.

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