P3 Differentiation Help Please!!

Hi, if anyone can help with any of the following I'd be very grateful:

1. Differentiate with respect to x:

(a) sinx/x
(b) ln (1/x^2 +9)

2. Given that y=cos2x+sinx, 0<x<2pi,

(a) find, in terms of pi, the values of x for which y=0
(b) find, to 2 decimal places the values of x for which dy/dx=0

3. Given that 2x^2 +6x+5 = P(x+Q)^2 + R, find the values of P, Q, R

Thank you so much for any help.

Reply 1

username9816

StuartYates

1. Differentiate with respect to x:
(a) sinx/x
(b) ln (1/x^2 + 9)

2. Given that y = cos2x + sinx, 0 < x < 2pi,
(a) find, in terms of pi, the values of x for which y = 0
(b) find, to 2 decimal places the values of x for which dy/dx = 0

3. Given that 2x^2 + 6x + 5 = P(x + Q)^2 + R, find the values of P, Q, R

1.) a.) Let y = sinx/x
Let u = sinx --> du/dx = cosx
Let v = x --> dv/dx = 1
d/dx [sinx/x] = dy/dx = [v(du/dx) - u(dv/dx)]/v^2
--> d/dx [sinx/x] = (xcosx - sinx)/(x^2)

b.) Let y = ln[1/(x^2 + 9)]
Let u = 1/(x^2 + 9) --> u = (x^2 + 9)^(-1)
Let z = x^2 + 9 --> dz/dx = 2x
Hence: u = 1/z = z^-1 --> du/dz = -z^-2 = -1/z^2
Hence: du/dx = du/dz * dz/dx = -1/(x^2 + 9)^2 * 2x = -2x/(x^2 + 9)^2
y = lnu --> dy/du = 1/u
Hence: d/dx [ln{1/(x^2 + 9)] = dy/dx = dy/du * du/dx = (x^2 + 9) * [-2x/(x^2 + 9)^2]
--> d/dx [ln{1/(x^2 + 9)] = -2x/(x^2 + 9)

2.) a.) For y = 0:
--> 0 = cos2x + sinx
--> 0 = 1 - 2sin^2x + sinx
--> 2sin^2x - sinx - 1 = 0
--> (2sinx + 1)(sinx - 1) = 0
--> sinx = -1/2, sinx = 1
In the required interval:
--> When sinx = -1/2: x = 7Pi/6, 11Pi/6.
--> When sinx = 1: x = Pi/2.
Hence for y = 0: x = Pi/2, 7Pi/6, 11Pi/6 (Rads)

b.) y = cos2x + sinx
--> dy/dx = cosx - 2sin2x
When dy/dx = 0:
--> 2sin2x = cosx
--> 2(2sinxcosx) = cosx
--> 4sinxcosx - cosx = 0
--> cosx(4sinx - 1) = 0
--> cosx = 0, sinx = 1/4.
In the required interval:
--> cosx = 0: x = Pi/2, 3Pi/2.
--> sinx = 1/4: x = 0.25, 2.89
Hence for dy/dx = 0: x = 0.25, 1.57, 2.89, 4.71 (Rads)

3.) 2x^2 + 6x + 5 = 2[x^2 + 3x + 2.5] = 2[(x + 1.5)^2 + 0.25] = 2(x + 1.5)^2 + 0.5
Hence: P = 2, Q = 1.5 and R = 0.5.

Reply 2

StuartYates

Thank you so much, that helps a lot; in understand the rest of the questions (bar one)now too!!

If you could just give me some help with it I'd be ever so grateful:

Given that y=x^x, by taking logarithms show that dy/dx = x^x(1+lnx)

Thank you so much again!!

Reply 3

username9816

StuartYates

Given that y = x^x, by taking logarithms show that dy/dx = x^x(1 + lnx)

y = x^x
--> lny = xlnx
--> d/dx [lny] = d/dx [xlnx]
--> (1/y) * dy/dx = x(lnx)' + lnx(x)'
--> dy/dx = y[x/x + lnx]
--> dy/dx = y(1 + lnx)
--> dy/dx = x^x.(1 + lnx)

Reply 4

StuartYates

OK, one more little question I'm a little stuck with if I could get some help please:

A curve has equation y=1/(2x^2 + 6x +5)

a) find the coordinates of the stationary point of this curve, state whether it is max or min.

the tangent to the curve at the point with coordinates (-1, 1) cuts the x-axis at M and the y-axis at N.

b) Using calculus and showing your working calculate coordinates of the mispoint MN.

Thanks for any help

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