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?

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.

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.

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.

Thank you very much for this valuable guidance.

where's my senpaii reiny?

1. Your reasoning is partially correct. When you want to determine whether a critical point corresponds to a maximum or minimum value, you typically analyze the sign of the second derivative. However, your reasoning for checking the sign of the numerator of the derivative is not entirely valid.

To determine the sign of the second derivative, you need to analyze the numerator specifically, not just the whole derivative expression. In your equation (A), the numerator is (1+tanx)cosx-sinx.sec^2(x). To check its sign, you would simplify it further and analyze when it is positive or negative.

2. If the sign of the second derivative is negative for a particular value of x, it will correspond to a maximum point. Conversely, if the sign of the second derivative is positive for a particular value of x, it will correspond to a minimum point. This is based on the Second Derivative Test, which states that if the second derivative is negative at a critical point, it is a local maximum; if the second derivative is positive at a critical point, it is a local minimum.

In summary, to determine whether a critical point corresponds to a maximum or minimum value:
- Find the second derivative.
- Analyze the sign of the second derivative.
- If the sign is negative, it corresponds to a maximum.
- If the sign is positive, it corresponds to a minimum.