Partial Derivatives MCQ Quiz in বাংলা - Objective Question with Answer for Partial Derivatives - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Jul 3, 2025
Latest Partial Derivatives MCQ Objective Questions
Partial Derivatives Question 1:
যদি y = log sin x হয়, তাহলে \(\frac{dy}{dx}\) হবে
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
Partial Derivatives Question 3 Detailed Solution
Download Solution PDFধারণা:
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?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
Partial Derivatives Question 6 Detailed Solution
\(f = {y^x}\)
ln f = x lny
⇒ \(\frac{1}{f}\frac{{df}}{{dy}} = \frac{x}{y}\)
⇒ \(\frac{{\partial f}}{{\partial y}} = {y^x}\left( {\frac{x}{y}} \right) = {y^{x - 1}}.x\)
⇒ \(\frac{{{\partial ^2}f}}{{\partial x\;\partial y}} = \frac{\partial }{{\partial x}}\left( {{y^{x - 1}}.x} \right)\)
\( = {y^{x - 1}} + x{y^{x - 1}}lny\)
\(= {1^{\left( {2 - 1} \right)}} + \left[ {2 \times {1^{\left( {2 - 1} \right)}}\ln \left( 1 \right)} \right] = 1\)