P3 Differentiation Watch

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Aristotle
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#1
Report Thread starter 13 years ago
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Find dy/dx for each of these relations:

1) x^2 + 2xy + 3y^2 = 6.

2) (e^x)(lny) = x.

3) x = asect, y = atant.

4) x = sin^2 t, y = costsint.

5) Find equations of the tangents to the curve with equation y = 2^x at the points P and Q where x=2 and x=5 respectively. These tangents meet at the point R. Find the x-coordinate of R.

6) Find an equation of the tangent at (2,1) to the curve with equation

y(x + y)^2 = 3(x^3 - 5).

7) Find dy/dx in terms of t for the curve with parametric equations

x = (t)/(1-t)

y = (t^2)/(1-t)

Deduce the equation of the normal at the point where t=0.5

8) Differentiate with respect to x:

a) 10^x.
b) 2^(x^2).
c) 5^(-x)

Answers:

1) -(x+y)/(x+3y).
2) y(e^-x - lny).
3) cosect.
4) cot2t.
5) (38ln2-7)/7ln2.
6) y=2x-3.
7) 2t-t^2, 6y+8x=11.
8a) (10^x)ln10
8b) 2x[2^(x^2)]ln2.
8c) (-5^-x)ln5.

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Aristotle
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#2
Report Thread starter 13 years ago
#2
Anybody?
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C4>O7
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#3
Report 13 years ago
#3
1. x^2 + 2xy + 3y^2 = 6.
Diff. wrt x:
2x+2x(dy/dx)+2y+6y(dy/dx)=0
dy/dx(2x+6y)+2y+2x=0
dy/dx=-(x+y)/(x+3y)
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C4>O7
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#4
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#4
e^xlny=x
Diff. wrt x:
e^x(1/y)(dy/dx)+e^xlny=1
dy/dx=y(1-e^xlny)/e^x = y(e^-x-lny)
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Nima
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#5
Report 13 years ago
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Find dy/dx for each of these relations:

1) x^2 + 2xy + 3y^2 = 6.
-> 2x + 2xdy/dx + 2y + 6ydy/dx = 0
-> dy/dx(2x + 6y) = -2(x + y)
-> dy/dx.2(x + 3y) = -2(x + y)
-> dy/dx.(x + 3y) = -(x + y)
-> dy/dx = -(x + y)/(x + 3y)

2) (e^x)(lny) = x.
-> (e^x).(1/y).(dy/dx) = 1
-> dy/dx.(e^x) = y
-> dy/dx = y/(e^x)

3) x = asect, y = atant.
-> dx/dt = asect.tant, dy/dt = asec^2t
dy/dx = dy/dt * dt/dx = asec^2t * 1/(asect.tant) = sect/tant = 1/(tant.cost) = 1/sint = cosect

4) x = sin^2t, y = costsint.
x = 1/2(1 - cos2t) -> dx/dt = (1/2)(2sin2t) = sin2t
y = cost.sint -> dy/dt = cos^2t - sin^2t = cos2t
dy/dx = dy/dt * dt/dx = cos2t/sin2t = 1/tan2t = cot(2t)

5) Find equations of the tangents to the curve with equation y = 2^x at the points P and Q where x = 2 and x = 5 respectively. These tangents meet at the point R. Find the x-coordinate of R.

y = 2^x -> lny = ln2^x = xln2 -> x = lny/ln2 = (1/ln2)(lny)
-> dx/dy = (1/ln2)(1/y) = 1/(yln2) -> dy/dx = yln2

At P, x = 2: -> y = 2^2 = 4 -> P: (2, 4).
At Q, x = 5: -> y = 2^5 = 32 -> Q: (5, 32).

At P: dy/dx = 4ln2 = ln16

-> Eq. Of Tangent at P: y - 4 = ln16(x - 2)
-> y - 4 = (ln16)x - 2ln16
-> y = (ln16)x + 4 - ln256

At Q: dy/dx = 32ln2
-> Eq. Of Tangent at Q: y - 32 = 32ln2(x - 5)
-> y - 32 = (32ln2)x - 160ln2
-> y = (32ln2)x + 32 - 160ln2

At R: (32ln2)x + 32 - 160ln2 = (ln16)x + 4 - ln256
-> (32ln2)x - (ln16)x = 160ln2 - ln256 - 28
-> (32ln2)x - (4ln2)x = 160ln2 - 8ln2 - 28
-> (28ln2)x = 152ln2 - 28
-> x = 38/7 - 1/(ln2) = (38ln2)/(7ln2) - 7/(7ln2)
-> x = (38ln2 - 7)/(7ln2)

6) Find an equation of the tangent at (2,1) to the curve with equation
y(x + y)^2 = 3(x^3 - 5).

->y[x^2 + 2xy + y^2] = 3x^3 - 15
-> yx^2 + 2xy^2 + y^3 = 3x^3 - 15
-> x^2(dy/dx) + 2yx + (2x)(2y)(dy/dx) + 2y^2 + 3y^2(dy/dx) = 9x^2
-> dy/dx[x^2 + 4xy + 3y^2] = 9x^2 - 2y^2 - 2yx
-> dy/dx = (9x^2 - 2y^2 - 2yx)/(x^2 + 4xy + 3y^2)

At (2, 1) -> dy/dx = (36 - 2 - 4)/(4 + 8 + 3) = 30/15 = 2

-> Eq. Of Tangent At (2, 1): y - 1 = 2(x - 2)
-> y - 1 = 2x - 4
-> y = 2x - 3

7) Find dy/dx in terms of t for the curve with parametric equations
x = (t)/(1-t) = t.(1 - t)^(-1)
y = (t^2)/(1-t) = t^2.(1 - t)^(-1)

dx/dt = t[(-1)(-1)(1 - t)^(-2)] + (1 - t)^(-1) = t(1 - t)^(-2) + (1 - t)^(-1)
= t/(1 - t)^2 + 1/(1 - t) = [t + (1 - t)]/(1 - t)^2 = 1/(1 - t)^2

dy/dt = t^2[(-1)(-1)(1 - t)^(-2)] + 2t(1 - t)^(-1) = t^2(1 - t)^(-2) + 2t(1 - t)^(-1) = t^2/(1 - t)^2 + 2t/(1 - t) = [t^2 + 2t(1 - t)]/(1 - t)^2 = [t^2 + 2t - 2t^2]/(1 - t)^2 = (2t - t^2)/(1 - t)^2 = t(2 - t)/(1 - t)^2

dy/dx = dy/dt * dt/dx = t(2 - t)/(1 - t)^2 * (1 - t)^2 = t(2 - t)

Deduce the equation of the normal at the point where t = 0.5.

Where t = 0.5:
-> x = 0.5/(1 - 0.5) = 0.5/0.5 = 1
-> y = (0.5)^2/(1 - 0.5) = (0.5)^2/(0.5) = 0.5

dy/dx = 0.5(2 - 0.5) = 0.5 * 1.5 = 3/4

-> Eq. Of Tangent at t = 5: y - 0.5 = (3/4)[x - 1]
-> y - 0.5 = 3x/4 - 3/4
-> y = 3x/4 - 1/4

8) Differentiate with respect to x:

a) 10^x.
Let y = 10^x -> lny = ln(10^x) = xln10 -> x = lny/ln10 -> dx/dy = (1/ln10)*(1/y) = 1/(yln10) -> dy/dx = yln10 = 10^x.(ln10)

-> d/dx[10^x] = 10^x.(ln10)

b) 2^(x^2).
Let y = 2^(x^2) -> lny = ln[2^(x^2)] = x^2ln2 -> x^2 = lny/ln2 -> 2x(dx/dy) = 1/(yln2) -> dx/dy = 1/[2xyln2] -> dy/dx = 2xyln2 = (2x).2^(x^2).ln2

-> d/dx[2^(x^2)] = (ln2)(2x)[2^(x^2)]

c) 5^(-x)
Let y = 5^(-x) -> lny = ln[5^(-x)] = -xln5 -> x = -lny/ln5 -> dx/dy = -1/(yln5) -> dy/dx = -yln5 = -ln5[5^(-x)] = -ln5/(5^x)

-> d/dx[5^(-x)] = -ln5/(5^x)
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C4>O7
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#6
Report 13 years ago
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3. x = asect, y = atant.
dy/dx=(dy/dt)/(dx/dt)=(asec²t)/(asect.tant)= cosect
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mathsforsaken
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#7
Report 13 years ago
#7
Lol, that's some excellent maths Nima.
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The Hunter
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#8
Report 13 years ago
#8
any help on my thing about opposite corners here, brothers?
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