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Differentiation problem-1

Posted: Fri Apr 27, 2012 10:48 am
by sm.joty
১.$x\sqrt(1+y)+y\sqrt(1+x)=0$
তাহলে, $\frac{dy}{dx}$ কে $x$ এর মাধ্যমে প্রকাশ করো। :)

২.$(cos x)^y=(sin y)^x$
$\frac{dy}{dx}= ?$

৩. $ln(x^{n}y^{n})=x^{n}+y^{n}$
$\frac{dy}{dx}= ?$

৪.$xy^{n}=yx^{n}$
$\frac{dy}{dx}= ?$

8-)

Re: Differentiation problem-1

Posted: Fri Apr 27, 2012 8:47 pm
by nafistiham
are these text book problems :?

Re: Differentiation problem-1

Posted: Tue May 01, 2012 11:24 am
by sm.joty
nafistiham wrote:are these text book problems :?
hmm...... but these are interesting. (actually all problems are from BUET admission test.) :D

Re: Differentiation problem-1

Posted: Tue May 01, 2012 2:12 pm
by Nadim Ul Abrar
Chain rule + trick $\frac{d}{dx} (f(x).g(x))$

we may take $ln$ in both side for some problems .

Re: Differentiation problem-1

Posted: Sun May 06, 2012 10:58 am
by sm.joty
What is the chain rules ??? I don't know actually. :?

Re: Differentiation problem-1

Posted: Sun May 06, 2012 12:34 pm
by shehab ahmed
if y and z are two functions of x then
$\frac {dz}{dx}=\frac {dy}{dx} \frac {dz}{dy}$
একে আরো বিস্তৃত করা যায়।এটাই chain rule

Re: Differentiation problem-1

Posted: Sun May 06, 2012 12:52 pm
by sm.joty
shehab ahmed wrote:if y and z are two functions of x then
$\frac {dz}{dx}=\frac {dy}{dx} \frac {dz}{dy}$
একে আরো বিস্তৃত করা যায়।এটাই chain rule
ও আচ্ছা সংযোজিত ফাংশনের নিয়মটাই Chain rules । হুম বুঝতে পারছি। ধন্যবাদ। :)