Partial Derivatives MCQ Quiz in বাংলা - Objective Question with Answer for Partial Derivatives - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Jul 3, 2025

পাওয়া Partial Derivatives उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Partial Derivatives MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Partial Derivatives MCQ Objective Questions

Partial Derivatives Question 1:

যদি y = log sin x হয়, তাহলে \(\frac{dy}{dx}\) হবে

  1. \(\frac{1}{sin~x} cos~x\)
  2. tan x
  3. \(\frac{1}{sin~x}\)
  4. log cos x

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{sin~x} cos~x\)

Partial Derivatives Question 1 Detailed Solution

ধারণা:

অবকলন শৃঙ্খল নিয়ম বলে যে, যদি y = f(u) এবং u = g(x) উভয়ই অকলনযোগ্য  অপেক্ষক হয়, তাহলে:

\(\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\)

\(\frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x},\;for\;x > 0\)

\(\frac{{d\left( {\sin x} \right)}}{{dx}} = \; cosx\)

গণনা:

প্রদত্ত: y = log sinx

ধরি sin x = u

⇒ y = log u

\(\frac{d}{{dx}}\left( {\log u} \right) = \frac{1}{{u}}\frac{d}{{dx}}\left( {u} \right) \)

\(= \frac{1}{{\sin x}}\left( { cos x} \right) \)

সুতরাং, \(\frac{dy}{dx}\) এর মান হবে \(\frac{1}{sin~x} cos~x\)

Partial Derivatives Question 2:

ধরি, f = yx , x = 2, y = 1 হলে \(\frac{{{\partial ^2}f}}{{\partial x\partial y}}\) এর মান কত হবে?

  1. 0
  2. ln 2
  3. 1
  4. \(\frac{1}{{\ln 2}}\)

Answer (Detailed Solution Below)

Option 3 : 1

Partial Derivatives Question 2 Detailed Solution

\(f = {y^x}\)

ln f = x lny

\(\frac{1}{f}\frac{{df}}{{dy}} = \frac{x}{y}\)

\(\frac{{\partial f}}{{\partial y}} = {y^x}\left( {\frac{x}{y}} \right) = {y^{x - 1}}.x\)

\(\frac{{{\partial ^2}f}}{{\partial x\;\partial y}} = \frac{\partial }{{\partial x}}\left( {{y^{x - 1}}.x} \right)\)

\( = {y^{x - 1}} + x{y^{x - 1}}lny\)

\(= {1^{\left( {2 - 1} \right)}} + \left[ {2 \times {1^{\left( {2 - 1} \right)}}\ln \left( 1 \right)} \right] = 1\)

Top Partial Derivatives MCQ Objective Questions

যদি y = log sin x হয়, তাহলে \(\frac{dy}{dx}\) হবে

  1. \(\frac{1}{sin~x} cos~x\)
  2. tan x
  3. \(\frac{1}{sin~x}\)
  4. log cos x

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{sin~x} cos~x\)

Partial Derivatives Question 3 Detailed Solution

Download Solution PDF

ধারণা:

অবকলন শৃঙ্খল নিয়ম বলে যে, যদি y = f(u) এবং u = g(x) উভয়ই অকলনযোগ্য  অপেক্ষক হয়, তাহলে:

\(\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\)

\(\frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x},\;for\;x > 0\)

\(\frac{{d\left( {\sin x} \right)}}{{dx}} = \; cosx\)

গণনা:

প্রদত্ত: y = log sinx

ধরি sin x = u

⇒ y = log u

\(\frac{d}{{dx}}\left( {\log u} \right) = \frac{1}{{u}}\frac{d}{{dx}}\left( {u} \right) \)

\(= \frac{1}{{\sin x}}\left( { cos x} \right) \)

সুতরাং, \(\frac{dy}{dx}\) এর মান হবে \(\frac{1}{sin~x} cos~x\)

ধরি, f = yx , x = 2, y = 1 হলে \(\frac{{{\partial ^2}f}}{{\partial x\partial y}}\) এর মান কত হবে?

  1. 0
  2. ln 2
  3. 1
  4. \(\frac{1}{{\ln 2}}\)

Answer (Detailed Solution Below)

Option 3 : 1

Partial Derivatives Question 4 Detailed Solution

Download Solution PDF

\(f = {y^x}\)

ln f = x lny

\(\frac{1}{f}\frac{{df}}{{dy}} = \frac{x}{y}\)

\(\frac{{\partial f}}{{\partial y}} = {y^x}\left( {\frac{x}{y}} \right) = {y^{x - 1}}.x\)

\(\frac{{{\partial ^2}f}}{{\partial x\;\partial y}} = \frac{\partial }{{\partial x}}\left( {{y^{x - 1}}.x} \right)\)

\( = {y^{x - 1}} + x{y^{x - 1}}lny\)

\(= {1^{\left( {2 - 1} \right)}} + \left[ {2 \times {1^{\left( {2 - 1} \right)}}\ln \left( 1 \right)} \right] = 1\)

Partial Derivatives Question 5:

যদি y = log sin x হয়, তাহলে \(\frac{dy}{dx}\) হবে

  1. \(\frac{1}{sin~x} cos~x\)
  2. tan x
  3. \(\frac{1}{sin~x}\)
  4. log cos x

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{sin~x} cos~x\)

Partial Derivatives Question 5 Detailed Solution

ধারণা:

অবকলন শৃঙ্খল নিয়ম বলে যে, যদি y = f(u) এবং u = g(x) উভয়ই অকলনযোগ্য  অপেক্ষক হয়, তাহলে:

\(\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\)

\(\frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x},\;for\;x > 0\)

\(\frac{{d\left( {\sin x} \right)}}{{dx}} = \; cosx\)

গণনা:

প্রদত্ত: y = log sinx

ধরি sin x = u

⇒ y = log u

\(\frac{d}{{dx}}\left( {\log u} \right) = \frac{1}{{u}}\frac{d}{{dx}}\left( {u} \right) \)

\(= \frac{1}{{\sin x}}\left( { cos x} \right) \)

সুতরাং, \(\frac{dy}{dx}\) এর মান হবে \(\frac{1}{sin~x} cos~x\)

Partial Derivatives Question 6:

ধরি, f = yx , x = 2, y = 1 হলে \(\frac{{{\partial ^2}f}}{{\partial x\partial y}}\) এর মান কত হবে?

  1. 0
  2. ln 2
  3. 1
  4. \(\frac{1}{{\ln 2}}\)

Answer (Detailed Solution Below)

Option 3 : 1

Partial Derivatives Question 6 Detailed Solution

\(f = {y^x}\)

ln f = x lny

\(\frac{1}{f}\frac{{df}}{{dy}} = \frac{x}{y}\)

\(\frac{{\partial f}}{{\partial y}} = {y^x}\left( {\frac{x}{y}} \right) = {y^{x - 1}}.x\)

\(\frac{{{\partial ^2}f}}{{\partial x\;\partial y}} = \frac{\partial }{{\partial x}}\left( {{y^{x - 1}}.x} \right)\)

\( = {y^{x - 1}} + x{y^{x - 1}}lny\)

\(= {1^{\left( {2 - 1} \right)}} + \left[ {2 \times {1^{\left( {2 - 1} \right)}}\ln \left( 1 \right)} \right] = 1\)

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