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Math Help - Calculus: Chain Rules and Implicit Differentiation

  1. #1
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    Calculus: Chain Rules and Implicit Differentiation

    I've actually already done the first 2 problems I'm posting, but I'd like to see how someone else works them out because I'm not sure about my work. Also, I definitely need to see how to work out the last two for I'm also not sure about them. You'll definitely be repped (thanked).

    1. Use the quotient rule to find the derivative of

    -8sin(x)+9/7x^6-7
    You do not need to expand out your answer.

    2. a)Find the derivative of: 6e-4xcos(9x). [Hint: use product rule and chain rule!]

    b) find the equation of the tangent line to the curve at x=0. Write your answer in mx+b format.

    3. a)Given the equation below, find dydx.

    10x^10+8x30y+y^2=19b) find the equation of the tangent line to the curve at (1, 1). Write your answer in mx+b format

    4. A fence 24 feet tall runs parallel to a tall building at a distance of 6 ft from the building.
    / | |
    / | |
    / | |
    / 24ft| | (bad pic, sorry)
    /______|___6ft_|

    We wish to find the length of the shortest ladder that will reach from the ground over the fence to the wall of the building.

    [A] First, find a formula for the length of the ladder in terms of θ. (Hint: split the ladder into 2 parts.)

    L(theta) = ?

    [b] Now, find the derivative, L'(θ).

    L'(theta) = ?

    [C] Once you find the value of θ that makes L'(θ)=0, substitute that into your original function to find the length of the shortest ladder. (Give your answer accurate to 5 decimal places.)

    L(theta(min)) is about ____?___ ft
    Last edited by tyuolio; May 21st 2007 at 06:42 PM.
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  2. #2
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    Quote Originally Posted by tyuolio View Post
    I

    1. Use the quotient rule to find the derivative of

    -8sin(x)+9/7x^6-7
    You do not need to expand out your answer.
    y=\frac{-8\sin x + 9 }{7x^6 - 7}

    y' = \frac{(-8\sin x+9)'(7x^6-7)-(-8\sin x + 9)(7x^6-7)'}{(7x^6-7)^2}

     = \frac{(-8\cos x)(7x^6-7)- (-8\sin x +9)(42x^5)}{(7x^6-7)^2}
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    Quote Originally Posted by tyuolio View Post

    2. a)Find the derivative of: 6e-4xcos(9x). [Hint: use product rule and chain rule!]
    y = 6e^{-4x}\cos 9x

    y' = (6e^{-4x})'\cos 9x + 6e^{-4x}(\cos 9x)'

    y' = -24e^{-4x}\cos 9x - 6e^{-4x} \cdot 9\sin 9x =-24e^{-4x}\cos 9x - 54 e^{-4x}\sin 9x
    b) find the equation of the tangent line to the curve at x=0. Write your answer in mx+b format.
    y'(0)=m = -24 e^0 \cos 0 - 6\cdot e^0 \cdot 0 = -24
    y(0) = 6e^0 \cos 0 = 6

    y-y(0) = y'(0)(x - 0) \to y-6 = -24x \to y = -24 x + 6
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    Quote Originally Posted by tyuolio View Post

    3. a)Given the equation below, find dydx.
    10x^10+8x30y+y^2=19b) find the equation of the tangent line to the curve at (1, 1). Write your answer in mx+b format
    10x^{10}+8x\cdot 30y+y^2=19

    100x^9 + \frac{d (8x)}{dx} \cdot 30 y + 8x \cdot \frac{dy}{dx} + 2y\frac{dy}{dx} = 0

    100x^9 + 240 y + 8x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0

     \frac{dy}{dx} \big( 8x + 2y \big) = - 100 x^9 - 240y

    \frac{dy}{dx} = \frac{-100 x^9 - 240 y}{8x+2y}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tyuolio View Post
    For this first one, I am getting y' = (-100x^9-240y)/(2y+240x). Can someone check this and see if they gett the same thing? It be very appreciative!
    And then in b, I got an answer also, but it depends if the first one is right.

    1. a)Given the equation below, find dydx.
    10x^10+8x^30*y+y^2=19
    b) find the equation of the tangent line to the curve at (1, 1). Write your answer in mx+b format
    we want to differentiate implicitly with respect to x. whenever we find the derivative of some x term we add dx/dx (that is, we took the derivative of x with respect x, but since derivative notations can function as fractions, we just cancel the dx's and get 1, so we don't write it). whenever we differentiate some y term, we attach dy/dx to it. note that if we are differentiating a product of x and y terms, we must use the product rule.

    (a)
    10x^10 + 8x^30*y + y^2 = 19 ......differentiating implicity w.r.t x, we get:

    100x^9 + 240x^29*y + 8x^30 dy/dx + 2y dy/dx = 0
    => 8x^30 dy/dx + 2y dy/dx = - 100x^9 - 240x^29*y
    => dy/dx(8x^30 + 2y) = - 100x^9 - 240x^29*y ...........factored out the dy/dx
    => dy/dx = (- 100x^9 - 240x^29*y)/(8x^30 + 2y)

    so yeah, i got the same as you did (you should really use parenthesis to make what you are saying clear)

    (b) at (1,1)
    dy/dx = (- 100(1)^9 - 240(1)^29*(1))/(8(1)^30 + 2(1))
    ........= (-100 - 240)/(8 + 2)
    ........= -34

    can you take it from here?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tyuolio View Post
    2. A fence 24 feet tall runs parallel to a tall building at a distance of 6 ft from the building.

    We wish to find the length of the shortest ladder that will reach from the ground over the fence to the wall of the building.
    [A] First, find a formula for the length of the ladder in terms of θ. (Hint: split the ladder into 2 parts.)
    L(theta) = ?
    [b] Now, find the derivative, L'(θ).
    L'(theta) = ?
    [C] Once you find the value of θ that makes L'(θ)=0, substitute that into your original function to find the length of the shortest ladder. (Give your answer accurate to 5 decimal places.)
    L(theta(min)) is about ____?___ ft
    Now i could sware i've seen this question posted around here some time ago. but nevermind.

    For problems like these, ALWAYS DRAW A DIAGRAM. see the diagram below:

    I labeled the appropriate angle t for theta. Note that we form two similar triagles, i called one A and one B. The hypotenuse of A is a and the hypotenuse of B is b. so we have the length of the ladder as a + b.

    usinf trig ratios we will realize that:

    in triagle A, the side opposite to t is 24, therefore we have:

    sin(t) = 24/a
    => a = 24/sin(t)

    in triangle B, the base is 6, therefore we have:

    cos(t) = 6/b
    => b = 6/cos(t)

    therefore, the length of the ladder is given by:

    L(t) = a + b = 24/sin(t) + 6/cos(t)

    i think you can take it from here
    Attached Thumbnails Attached Thumbnails Calculus: Chain Rules and Implicit Differentiation-ladder.gif  
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    Now i could sware i've seen this question posted around here some time ago. but nevermind.

    For problems like these, ALWAYS DRAW A DIAGRAM. see the diagram below:

    I labeled the appropriate angle t for theta. Note that we form two similar triagles, i called one A and one B. The hypotenuse of A is a and the hypotenuse of B is b. so we have the length of the ladder as a + b.

    usinf trig ratios we will realize that:

    in triagle A, the side opposite to t is 24, therefore we have:

    sin(t) = 24/a
    => a = 24/sin(t)

    in triangle B, the base is 6, therefore we have:

    cos(t) = 6/b
    => b = 6/cos(t)

    therefore, the length of the ladder is given by:

    L(t) = a + b = 24/sin(t) + 6/cos(t)

    i think you can take it from here
    awesome. thank you very much jhevon.
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