📝 Derivative of sin x and cos x (16 MCQs)
📖 From Calculus • 3. The Derivation • 16 questions available
What is Derivative of sin x and cos x?
Definition:
The derivative of is , and the derivative of is , reflecting the periodic nature of trigonometric functions and their phase shifts in rate of change across the unit circle.
Example:
For , the derivative is , applying both rules simultaneously.
Reason:
These fundamental trigonometric derivatives are indispensable in modeling oscillatory phenomena, wave mechanics, and circular motion, forming the backbone of calculus applications in physics and engineering disciplines.
📝 All Derivative of sin x and cos x MCQs
Q1. What is the derivative of \\\sin x\ with respect to \x\?
📖 Explanation: The derivative of \\\sin x\ follows the basic rule \\\frac{d}{dx}\\sin x = \\cos x\. Among the listed choices only \\\cos x\ correctly represents this rate of change, while the others either have the wrong sign or correspond to a different trigonometric function.
Q2. Evaluate \\\frac{d}{dx}\\cos x\ at \x = \\frac{\\pi}{3}\.
📖 Explanation: Differentiating \\\cos x\ gives \-\\sin x\. Substituting \x=\\frac{\\pi}{3}\ yields \-\\sin\\frac{\\pi}{3}\ = \-\\frac{\\sqrt{3}}{2}\. Hence the correct choice is \-\\sqrt{3}/2\.
Q3. If \f(x)=\\sin x+\\cos x\, what is \f'(\\frac{\\pi}{4})\?
📖 Explanation: The derivative of \f\ is \f'(x)=\\cos x-\\sin x\. At \x=\\frac{\\pi}{4}\ both \\\cos\ and \\\sin\ equal \\\frac{\\sqrt{2}}{2}\, so their difference is zero. Therefore the correct answer is 0.
Q4. Compare the instantaneous rates of change of \\\sin x\ and \\\cos x\ at \x = \\frac{\\pi}{6}\. Which is larger?
📖 Explanation: The rate of change of \\\sin x\ is \\\cos x\ = \\\frac{\\sqrt{3}}{2}\ ≈ 0.866, while the rate of \\\cos x\ is \-\\sin x\ = \-\\frac{1}{2}\ ≈ -0.5. Since 0.866 > -0.5, the derivative of \\\sin x\ is larger.
Q5. Find \\\frac{d}{dx}\\bigl(\\sin(3x^{2})\\bigr)\.
📖 Explanation: Applying the chain rule, the outer derivative of \\\sin u\ is \\\cos u\ and the inner derivative of \u=3x^{2}\ is \6x\. Multiplying gives \6x\\cos(3x^{2})\, which matches option B.
Q6. A shadow length is given by \s = 50\\cot\\theta\. If \\\theta\ increases at \0.02\ rad/s, what is \\\frac{ds}{dt}\ when \\\theta = \\frac{\\pi}{4}\?
📖 Explanation: First compute \\\frac{ds}{d\\theta}= -50\\csc^{2}\\theta\. At \\\theta=\\frac{\\pi}{4}\, \\\csc^{2}\\theta=2\, so \\\frac{ds}{d\\theta}= -100\. Multiplying by \d\\theta/dt=0.02\ gives \\\frac{ds}{dt}= -2\ ft/s.
Q7. What is the second derivative \\\frac{d^{2}}{dx^{2}}\\sin x\?
📖 Explanation: The first derivative of \\\sin x\ is \\\cos x\. Differentiating again yields \-\\sin x\. Thus the second derivative is \-\\sin x\, corresponding to option C.
Q8. Which of the following statements about the graph of \\\sin x\ is true?
📖 Explanation: The sine function reaches its maximum value of 1 at \x = \\frac{\\pi}{2}\ and then repeats every \2\\pi\. None of the other statements correctly describe the behavior of \\\sin x\.
Q9. Define \g(x)=\\frac{\\sin x}{x}\ for \x\\neq0\ and \g(0)=1\. What is \g'(0)\?
📖 Explanation: Using the limit definition, \g'(0)=\\lim_{h\\to0}\\frac{\\sin h/h-1}{h}=\\lim_{h\\to0}\\frac{\\sin h - h}{h^{2}}\. Expanding \\\sin h = h - h^{3}/6+\\dots\ gives a numerator of order \h^{3}\, so the quotient tends to 0. Hence \g'(0)=0\.
Q10. If \y=\\sin^{2}x\, which expression gives \\\frac{dy}{dx}\ and at which points in \[0,2\\pi]\ does the derivative equal zero?
📖 Explanation: Differentiating using the chain rule yields \\\frac{dy}{dx}=2\\sin x\\cos x=\\sin 2x\. This expression is zero whenever \\\sin x=0\ or \\\cos x=0\, giving the points \0,\\frac{\\pi}{2},\\pi,\\frac{3\\pi}{2},2\\pi\. Option A lists the full set correctly.
Q11. Evaluate \\\displaystyle\\lim_{h\\to0}\\frac{\\cos(x+h)-\\cos x}{h}\.
📖 Explanation: The limit matches the definition of the derivative of \\\cos x\. Since \\\frac{d}{dx}\\cos x = -\\sin x\, the limit equals \-\\sin x\, which is option A.
Q12. A particle has position \s(t)=\\cos t\. At which times in \[0,2\\pi]\ does its acceleration equal its position?
📖 Explanation: The acceleration is the second derivative: \s''(t)= -\\cos t\. Setting \s''(t)=s(t)\ gives \-\\cos t = \\cos t\ → \\\cos t = 0\. Within the interval this occurs at \t=\\frac{\\pi}{2}\ and \t=\\frac{3\\pi}{2}\.
Q13. To simplify the derivative of \h(x)=\\sin x\\cos x\, which trigonometric identity is most useful?
📖 Explanation: The product \\\sin x\\cos x\ can be rewritten using the double‑angle identity \\\sin 2x = 2\\sin x\\cos x\. This conversion reduces the product to a single sine function, making differentiation straightforward.
Q14. What is \\\frac{d}{dx}\\tan x\?
📖 Explanation: The derivative of the tangent function follows the rule \\\frac{d}{dx}\\tan x = \\sec^{2}x\. This result comes from differentiating \\\sin x/\\cos x\ or from the known identity, confirming option C.
Q15. When converting \\\frac{ds}{d\\theta}\ from radians to degrees, which of the following is true?
📖 Explanation: Conversion from radians to degrees multiplies the derivative by \\\pi/180\. This factor does not alter the sign of the original rate, so a negative derivative stays negative after conversion.
Q16. For \f(x)=\\sin x+\\cos 2x\, at which \x\ in \[0,2\\pi]\ does a local maximum occur?
📖 Explanation: Setting \f'(x)=\\cos x-2\\sin 2x=0\ leads to \\\sin x=\\frac{1}{4}\. Evaluating the second derivative \f''(x)=-\\sin x-4\\cos 2x\ at this solution gives a negative value, confirming a local maximum at \x=\\arcsin(1/4)\.)