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If cosy=xcos(a+y). And dx/dy =k/1+(x)2-2xcosa...... Then find value of k Now my main question is..For above que. We should diff. w.r.t "y" and we get k=sina.. ..And i see same que in diff book like this mainor.. .... If cosy=xcos(a+y). And dy/dx =k/1+(x)2-2xcosa...... Then find value of k ,and for this we should diff w.r.t "x"..And we get same ans. K=sina then for this que. We prove dy/dx=dx/dy... Thanks

3 years ago


Answers : (1)


may be the question is wrong or may be the value ofdy/dx ...or dx/dy are equal to 1 at k = sina

2 years ago

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2^x=3^y=6^z , then values of x,y&z....
But in general we cannot identify wheter the problem is related to log or something else I am also getting the same answer Can u pl. Tell the correct answer
grenade 3 months ago
it might be the case that u had printing error while posting the answer else you just check out once againg and i will approach u back in few hours
grenade 3 months ago
if we have to prove this result any way then we could easily say that x=y=z=0 which will give us 1=1=1 SMASH THE APPROVE BUTTON IF IT HELPED U
Gman Namg 3 months ago
Prove that, If y = Sin -1 x, x ∈ (–1, 1) then dy/dx =1/√(1-x 2)
y = Sin -1 x ⇒ x = sin y dy/dx=cosy dy/dx=(dx/dy) -1 =1/cosy=1/ √(1-sin 2 y) =1/√(1-x 2 )
RAJU 28 days ago
y = Sin -1 x ⇒ x = sin y dy/dx=cosy dy/dx=(dx/dy) -1 =1/cosy=1/ √(1-sin 2 y) =1/√(1-x 2 )
Raghu Vamshi Hemadri 22 days ago
I m having problem in these two questions: 1) f(x)=x^3+6x^2+(9+2k)x+1 is increasing if k is 2)f(x)=cosx-2PX is monotonically decreasing for p:
HInt: 1. If we diffirenciate a cubic equation, we get the quadratic equation. Not for cubic polynomial to be increasing function, its derivative should be positive, i.e. quadratic polynomial...
Vijay Mukati 2 months ago
Give an example of a sequence {an} such that it passes three limit points.[Hint:Example of a sequence {an} passes two limits is {(-1)^n}]
First off, a limit point for a real number sequence is a real number that has infinitely many terms at any distance from it (i.e. in any epsilon neighbourhood). This is a relaxed condition...
mycroft holmes one month ago
1,-1,1,1,-1,1/2, 1,-1,1/3,.... i.e. the sequence given by a n = 1 if n=3k+1 = -1 if n = 3k+2 = 1/k if n = 3k has infinitely many terms that are in any neighbourhood of 1,-1, and 0. Hence...
mycroft holmes one month ago
can you please explain it......
MOHAMMAD AFROZ one month ago
2+7+14+23+.... then find the sum upto 30 terms
2, 7, 14, 23, 34 …........ is a series whose common difference is an AP (i.e 5, 7, 9, 11,….). S= 2 + 7 + 14 + 23 + ……………. A n-1 + A n S= 2 + 7 + 14+ 23+………………+A n-1 +A n Now, subtracting...
AUREA 4 months ago
it is 2 2 -2 + 3 2 -2 + 4 2 -2 till 31 2 -2 we will get is as (∑n 2 ) -1 * – 2(30) -1 * is taken because n 2 starts from 1 2+ 2 2 where ∑n 2 =n(n+1)(2n+1)/6 on substituting the values it...
grenade 4 months ago
putting the value n=30 ∑n 2 =9455 2∑n=930 finally the sum comes to be S=9455+930-30 ,i.e, S=10355 approve if the solution is helpful ; )
AUREA 4 months ago
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