# Calculus: Chain Rules and Implicit Differentiation

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• May 21st 2007, 03:35 PM
tyuolio
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
• May 21st 2007, 06:44 PM
ThePerfectHacker
Quote:

Originally Posted by tyuolio
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.

$y=\frac{-8\sin x + 9 }{7x^6 - 7}$

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

$= \frac{(-8\cos x)(7x^6-7)- (-8\sin x +9)(42x^5)}{(7x^6-7)^2}$
• May 21st 2007, 06:50 PM
ThePerfectHacker
Quote:

Originally Posted by tyuolio

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

$y = 6e^{-4x}\cos 9x$

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

$y' = -24e^{-4x}\cos 9x - 6e^{-4x} \cdot 9\sin 9x =-24e^{-4x}\cos 9x - 54 e^{-4x}\sin 9x$
Quote:

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

$y-y(0) = y'(0)(x - 0) \to y-6 = -24x \to y = -24 x + 6$
• May 21st 2007, 06:56 PM
ThePerfectHacker
Quote:

Originally Posted by tyuolio

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

$10x^{10}+8x\cdot 30y+y^2=19$

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

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

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

$\frac{dy}{dx} = \frac{-100 x^9 - 240 y}{8x+2y}$
• May 21st 2007, 10:01 PM
Jhevon
Quote:

Originally Posted by tyuolio
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?
• May 21st 2007, 10:21 PM
Jhevon
Quote:

Originally Posted by tyuolio
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
• May 22nd 2007, 08:53 AM
tyuolio
Quote:

Originally Posted by Jhevon
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.