differentiate
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fast huh..^_^ ^_^ ^_^
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itta clear to h
3rd step
are clarity nai steps
Samaj gaya vo Kya h tube light thodi Purani hai
third step me simplify kiya h jitta common h sab ko eksaath nikaal k rhs me bhej diya h
ha
are Maine Dhayn se nai dekha tha to kuch samja nai
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Question ⤵️
if sqrt(y+x)+ sqrt(y-x)=c, show that dy/dx=y/x-sqrt(y2/x2-1).
the given equation is : y+x−−−−√+y−x−−−−√=c ..........(1)
differentiating wrt x:
12y+x√.[dydx+1]+12y−x√.[dydx−1]=0
[1y+x√+1y−x√].dydx+[1y+x√−1y−x√]
=0 [multipliyng eq by 2]
[y−x√+y+x√(y−x)(y+x)√].dydx [y−x√−y+x√(y−x)(y+x)√]=0
[y−x−−−−√+y+x−−−−√].dydx+[y−x−−−−√−y+x−−−−√]=0
[multiplying whole eq. by (y−x)(y+x)−−−−−−−−−−−√]
dydx=−y−x√−y+x√y−x√+y+x√dydx=−y−x√−y+x√y−x√+y+x√*y−x√−y+x√y−x√−y+x✓ ⎡
⎣rationalising the denominator⎤
⎦dydx=−(y−x√−y+x√)^2(y−x)−(y+x)
dydx=−y−x+y+x−2.y−x√.y+x√−2xdydx=2y−2(y−x)(y+x)√2x=y−(y−x)(y+x)√xdydx=yx−y2−x2√xdydx=yx−y2x2−1−−−−−√
which is the required result.
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