📝 Derivatives of inverse trig functions arcsin arccos arctan (18 MCQs)
📖 From Calculus • 4. Topics in Differentiation • 18 questions available
What is Derivatives of inverse trig functions arcsin arccos arctan?
Definition:
The derivatives of inverse trigonometric functions are algebraic expressions involving square roots. Specifically, , , and . These formulas are derived using implicit differentiation on their corresponding trigonometric identities and are valid within their respective domains.
Example:
Find the derivative of . Using the chain rule, . For , .
Reason:
These derivatives appear frequently in integration techniques and physics problems involving angles and circular motion. Knowing them allows for the direct differentiation of composite functions involving inverse trigonometric terms without reverting to implicit methods each time.
📝 All Derivatives of inverse trig functions arcsin arccos arctan MCQs
Q1. Given that \\\arcsin x\ is increasing on \(-1,1)\, what can be inferred about the sign of its derivative on that interval?
📖 Explanation: Because \\\arcsin x\ is strictly increasing throughout its domain, the slope of its tangent line must be positive at every interior point. Therefore the derivative \\\frac{d}{dx}\\arcsin x\ is positive for all \x\ in \(-1,1)\.
Q2. Suppose \g(x)=\\arctan(f(x))\ where \f\ is increasing and \f'(x)>0\. What can be said about \g'(x)\?
📖 Explanation: Both \f\ and \\\arctan\ are increasing functions, so their derivatives are positive. By the chain rule, \g'(x)=f'(x)/(1+f(x)^2)\; the denominator is always positive, thus the sign of \g'\ matches the sign of \f'\, which is positive.
Q3. If \h(x)=\\arccos x\ and \h'(c)=-\\frac{1}{\\sqrt{1-c^2}}\, what is the monotonicity of the inverse function \h^{-1}\ (i.e., restricted \\\cos x\)?
📖 Explanation: The derivative of \\\arccos x\ is negative on its domain, indicating that \\\arccos\ is decreasing. The derivative of the inverse function is the reciprocal, \(h^{-1})'(y)=1/h'(x)\, which inherits the negative sign, so the inverse is also decreasing.
Q4. Determine the monotonicity of \\\operatorname{arcsec} x\ on its domain.
📖 Explanation: The derivative of \\\operatorname{arcsec} x\ is \1/(|x|\\sqrt{x^2-1})\. For \x\\ge1\ this expression is positive, giving an increasing function; for \x\\le-1\ it is negative, giving a decreasing function. Hence the mixed monotonicity described.
Q5. For \p(x)=\\arcsin(x^3)\, what is the sign of \p'(0)\?
📖 Explanation: Using the chain rule, \p'(x)=\\frac{3x^2}{\\sqrt{1-x^6}}\. At \x=0\ the numerator is zero while the denominator is non‑zero, so the derivative equals zero. Thus the sign of \p'(0)\ is zero.
Q6. Suppose a function \f\ is invertible with \f'(a)=2\. The inverse derivative at \b=f(a)\ is \0.5\. If we define \g(x)=\\arctan(f(x))\, what is \g'(a)\?
📖 Explanation: Applying the chain rule, \g'(a)=\\frac{f'(a)}{1+f(a)^{2}}\. With \f'(a)=2\ and \f(a)=b\, this yields \g'(a)=2/(1+b^{2})\. The given value of the inverse derivative is consistent but not needed for this calculation.
Q7. Compare the derivatives of \\\arcsin x\ and \\\arccos x\. Which statement is true?
📖 Explanation: Differentiating each gives \\\frac{d}{dx}\\arcsin x=\\frac{1}{\\sqrt{1-x^{2}}}\ and \\\frac{d}{dx}\\arccos x=-\\frac{1}{\\sqrt{1-x^{2}}}\. The two formulas differ only by a sign, so the derivatives are exact negatives of each other.
Q8. At \x=0.5\, which inverse trigonometric function has the largest magnitude derivative?
📖 Explanation: For \x=0.5\, \\\arcsin' =1/\\sqrt{1-0.25}=1/\\sqrt{0.75}\\approx1.155\. \\\arccos'\ has the same magnitude. \\\arctan' =1/(1+0.25)=0.8\. \\\arcsec\ is undefined at 0.5. Thus the largest magnitude belongs to \\\arcsin\ (and equally to \\\arccos\).
Q9. Differentiate \h(x)=\\arcsin(\\sqrt{x})\.
📖 Explanation: Let \u=\\sqrt{x}\. Then \h'(x)=u' /\\sqrt{1-u^{2}}\ with \u'=1/(2\\sqrt{x})\ and \1-u^{2}=1-x\. Hence \h'(x)=\\frac{1}{2\\sqrt{x}\\sqrt{1-x}}\.
Q10. Which of the following functions is not differentiable at \x=0\?
📖 Explanation: The function \\\arcsec x\ is defined only for \|x|\\ge1\; it is undefined at \x=0\, so it cannot be differentiated there. All other listed inverse trigonometric functions are defined and differentiable at \0\.
Q11. Consider \k(x)=\\arctan\\left(\\frac{1}{x}\\right)\ for \x>0\. Find \k'(x)\.
📖 Explanation: Let \u=1/x\; then \u'=-1/x^{2}\. Using the chain rule, \k'(x)=u'/(1+u^{2}) = -\\frac{1}{x^{2}}\\big/\\left(1+\\frac{1}{x^{2}}\\right)= -\\frac{1}{x^{2}+1}\.
Q12. Explain why the derivative formula for \\\arcsin x\ involves \\\sqrt{1-x^{2}}\.
📖 Explanation: Implicitly differentiate \\\sin y = x\. Differentiating both sides gives \\\cos y\\,dy/dx = 1\. Using the identity \\\cos y = \\sqrt{1-\\sin^{2} y}=\\sqrt{1-x^{2}}\ yields \dy/dx = 1/\\sqrt{1-x^{2}}\. Hence the square‑root appears.
Q13. To integrate \\\frac{1}{\\sqrt{1-x^{2}}}\, which inverse trigonometric function's derivative matches the integrand?
📖 Explanation: The derivative of \\\arcsin x\ is exactly \1/\\sqrt{1-x^{2}}\. Recognizing this match allows us to write \\\int\\frac{1}{\\sqrt{1-x^{2}}}\\,dx = \\arcsin x + C\.
Q14. If a function \f\ is invertible and its derivative never zero, explain how monotonicity guarantees the existence of \f^{-1}\'s derivative and its sign.
📖 Explanation: When \f'\ never vanishes, \f\ is strictly monotonic, ensuring a one‑to‑one mapping and a differentiable inverse. The inverse derivative formula \(f^{-1})'(y)=1/f'(f^{-1}(y))\ shows that the sign of \(f^{-1})'\ matches the sign of \f'\; thus an increasing \f\ yields an increasing inverse, and a decreasing \f\ yields a decreasing inverse.
Q15. Apply the principle to find the derivative of \\\operatorname{arccsc} x\.
📖 Explanation: Let \y=\\operatorname{arccsc} x\ so \\\csc y = x\. Differentiating gives \-\\csc y\\cot y\\,dy/dx = 1\. Using \\\csc y = x\ and \\\cot y = \\sqrt{x^{2}-1}\, we obtain \dy/dx = -1/(|x|\\sqrt{x^{2}-1})\.
Q16. Which inverse trigonometric function has a vertical asymptote in its derivative?
📖 Explanation: The derivative of \\\arcsec x\ is \1/(|x|\\sqrt{x^{2}-1})\. As \x\ approaches \1\ or \-1\ from within the domain, the denominator tends to zero, causing the derivative to blow up and create a vertical asymptote.
Q17. Evaluate \\\lim_{x\\to1^{-}}\\frac{d}{dx}\\arccos x\.
📖 Explanation: The derivative of \\\arccos x\ is \-1/\\sqrt{1-x^{2}}\. As \x\ approaches \1\ from the left, the denominator \\\sqrt{1-x^{2}}\ tends to zero positive, making the fraction tend toward \-\\infty\.
Q18. What is the derivative of \\\arctan x\?
📖 Explanation: Differentiating \\\arctan x\ using the standard formula gives \\\frac{d}{dx}\\arctan x = \\frac{1}{1+x^{2}}\. The denominator is always positive, ensuring the derivative is well‑defined for all real \x\.