Question

Q: If y=sinx/(1+tanx), find value of x not greater than pi, corresponding to maxima or minima value of y. I have proceeded thus-
Equating dy/dx=0 we get{ (1+tanx)cosx-sinx.sec^2 x}/(1+tanx)^2=0……..(A)
Or cosx+sinx=sinx.sec^2 x or cosx= sinx.sec^2-sinx
Or cosx=sinx.tan^2 x=sinx.sin^2 x/cos^2 x
Or cos^3 x=sin^3 x hence cosx=sinx=pi/4. Since working out II DC is very tedious, I have reasoned that to check the sign of second DC, sign of Nr of dy/dx( i.e. A) can be checked since its denominator i.e. (1+tanx)^2 and its DC i.e. sec^2 x are always +ve. Hence,
DC of Nr of (A)= -sinx+cosx-{cosx.sec^2 x+sinx.2sec^2 x.tanx}
= -sinx+cosx-secx-2tanx/cos^2 x
At x=pi/4, it is = -2/sqrt2 + 1/sqrt2 - sqrt2- 2/2= -2sqrt2 + 1/sqrt2 -1
= (-3-sqrt2)/sqrt2 which is negative, hence x=pi/4 is maxima point. Answers are correct but,
1. Is my reasoning correct?
2. If sign of expression for II DC in general is –ve, but for a particular value of x its value is +ve, will it correspond to maxima or minima?

Answer 1

From what I can decipher from your cryptic use use of arcana (like A, II DC, Nr, etc.) your reasoning appears sound. I have a bit of trouble distinguishing between II DC and second DC. If they are the same, how can you check the sign of second DC without calculating it?

If y'(a)=0, and y' goes from positive to negative at x=a, then it is a maximum, since y has stopped rising it starts to fall there.

The sign of y" indicates concavity, so if y'=0 and y" < 0, it is a maximum.

Answer 2

Thanks for reply. Sorry for confusion. Second and II DC are same and only mean second diff. coeff. or derivative d^2y/dx^2. Nr is numerator and A is dy/dx as shown as equation(A)in the first line of answer.
What I meant was that for calculating second DC we have to apply quotient formula in which denominator and its DC will always be +ve. Can't the sign of rest of the expression determine the sign of the second DC, without the need for working it out fully?

For second part, I wondered if in some case I get an extrema at x= -5 by putting dy/dx=0 and second DC is -x, then is it max, or can it be a min as value of second DC at x=-5 will be +5.

Answer 3

Your function becomes less unwieldy if you note that if you divide top and bottom by cosx, you have

y = (1/2 sin2x)/(sinx+cosx)

All those nasty sec^2 and stuff go away and you have

2nd: wolframalpha.com is your friend. Enter

2nd derivative sinx/(1+tanx)

and it pops right up.

If, as you propose, y'(-5)=0 and y" = -x, then y" > 0 and so y is concave up, making it a minimum at x = -5.

Answer 4

Thank you very much for this valuable guidance.

Answer 5

where's my senpaii reiny?