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?

  1. 👍
  2. 👎
  3. 👁
  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.

    1. 👍
    2. 👎
  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.

    1. 👍
    2. 👎
  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: 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.

    1. 👍
    2. 👎
  4. Thank you very much for this valuable guidance.

    1. 👍
    2. 👎
  5. where's my senpaii reiny?

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. Math (Calculus AB)

    For x≠0, the slope of the tangent to y=xcosx equals zero whenever: (a) tanx=-x (b) tanx=1/x (c) sinx=x (d) cosx=x Please help. I have a final tomorrow and I am working diligently to understand every type of problem that may show

  2. Math

    How do I solve this? tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0 and sinX/cosX = 2 sin X Divide through by sinX: we have to check this later to see if allowed (ie sinX

  3. Calculus

    True or False: Consider the following statement: A differentiable function must have a relative minimum between any two relative maxima. Think about the First Derivative Test and decide if the statement is true or false. I want to

  4. Trig

    Verify the identity: tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx cosx - sinx/cosx =(2sinxcos^2 x - sinx)/cosx =sinx(2cos^2 x -1)/cosx = L.S. Q.E.D.

  1. college algebra

    Use the graph of the given function to find any relative maxima and relative minima. f(x) = x^3 - 3x^2 + 1

  2. Trigonometry.

    ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated

  3. physics

    A monochromatic light illuminates a double slits system with a slit separation, d =0.3 mm. The maximum of the first order occurs at y = 4.0 mm on a screen placed at 1.0 m from the slits. Find the: a)Wavelength b)Distance between

  4. Mathematics - Trigonometric Identities

    Prove: sinx + tanx = tanx (1 + cosx) What I have so far: LS: = sinx + tanx = sinx + (sinx / cosx) = (sinx) (cosx) + sinx / cos = tanx (cosx + sinx) I don't know what to do now

  1. Trigonometry

    Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do.

  2. TRIG

    There are two questions of my homework I'm having trouble with.. I think we are supposed to show how they are true. In other words, make one side look exactly like the other one by using the identities. 1.

  3. Precalculus

    Given sinx= -1/8 and tanx

  4. Math 12

    Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do.

You can view more similar questions or ask a new question.