large{ underline{ underline{ bigstar : : : : pmb{ sf{Question : : }}}}} odot : :

Question

large{  underline{  underline{  bigstar  :   :   :   :   pmb{  sf{Question  :  : }}}}}     odot  :   :   :   : ɪғ  :   :   sf{ {x}^{5}  +  {y}^{5}  =  {x}^{3}  {y}^{2} }, sʜᴏᴡ   :   : ᴛʜᴀᴛ  :   :   sf {  frac{dy}{dx}  =   frac{y}{x} }      ​

Camila · Answer

Appropriate Question  :-       rm : longmapsto :ɪғ  :  {(x + y)}^{5} =  {x}^{3} {y}^{2},  : sʜᴏᴡ  : ᴛʜᴀᴛ  :  dfrac{dy}{dx} =  dfrac{y}{x}       Formula Used  :-          red{  boxed{  sf{  :  log(xy) = logx + logy }}}      red{  boxed{  sf{  log( {x}^{y} )  :  =   : y  : logx}}}      red{  boxed{  sf{  dfrac{d}{dx}x :  =   : 1}}}      red{  boxed{  sf{  dfrac{d}{dx}logx :  =   :   frac{1}{x} }}}        Solution  :-       Given function is       rm : longmapsto : {(x + y)}^{5} =  {x}^{3} {y}^{2}      On taking log both sides, we get       rm : longmapsto :log {(x + y)}^{5} =log(  {x}^{3} {y}^{2} )     rm : longmapsto :5 log(x + y) =  log( {x}^{3} ) +  log( {y}^{2} )      rm : longmapsto :5 log(x + y) = 3 log( {x}) +  2log( {y} )      On differentiating both sides w. r. t. x, we get      rm : longmapsto : dfrac{d}{dx}5 log(x + y) =  dfrac{d}{dx}3 log( {x}) +  dfrac{d}{dx} 2log( {y} )      rm : longmapsto :5 dfrac{d}{dx} log(x + y) = 3 dfrac{d}{dx}log( {x}) +2  dfrac{d}{dx} log( {y} )      rm : longmapsto :5  times  dfrac{1}{x + y}  bigg(1 +  dfrac{dy}{dx}  bigg) =  dfrac{3}{x}  +  dfrac{2}{y} dfrac{dy}{dx}     rm : longmapsto : dfrac{5}{x + y}  bigg(1 +  dfrac{dy}{dx}  bigg) =  dfrac{3}{x}  +  dfrac{2}{y} dfrac{dy}{dx}     rm : longmapsto : dfrac{5}{x + y}  +  dfrac{5}{x + y}   dfrac{dy}{dx} =  dfrac{3}{x}  +  dfrac{2}{y} dfrac{dy}{dx}     rm : longmapsto :   dfrac{5}{x + y}   dfrac{dy}{dx}  -  dfrac{2}{y} dfrac{dy}{dx}=  dfrac{3}{x}   -  dfrac{5}{x + y}      rm : longmapsto :    bigg( dfrac{5}{x + y}   -  dfrac{2}{y}  bigg) dfrac{dy}{dx}=  dfrac{3}{x}   -  dfrac{5}{x + y}      rm : longmapsto :    bigg( dfrac{5y - 2x - 2y}{y(x + y)}    bigg) dfrac{dy}{dx}=  dfrac{3x + 3y - 5x}{x(x + y)}       rm : longmapsto :    bigg( dfrac{3y - 2x}{y(x + y)}    bigg) dfrac{dy}{dx}=  dfrac{3y - 2x}{x(x + y)}       rm : longmapsto : dfrac{1}{y} dfrac{dy}{dx} =  dfrac{1}{x}     bf implies  : dfrac{dy}{dx} =  dfrac{y}{x}         Short Cut Trick         boxed{ bf : longmapsto :ɪғ  :  {(x + y)}^{m + n} =  {x}^{m} {y}^{n},  : then :  dfrac{dy}{dx} =  dfrac{y}{x} }

Melanie

#include <stdio.h>
void main(){
int i=0;
if(printf(“%”,i)==0) {
printf(” if block executed\n”);
}

Change the following sentence into passive voice.
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1 thought on “<br />[tex] \large{ \underline{ \underline{ \bigstar \: \: \: \: \pmb{ \sf{Question \: : }}}}}[/tex]<br />[tex] \odot \: \:”

Appropriate Question :-

Formula Used :-

Solution :-

Short Cut Trick

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[tex] \large{ \underline{ \underline{ \bigstar \: \: \: \: \pmb{ \sf{Question \: : }}}}}[/tex][tex] \odot \: \: