implicit differentiation

Hi guys,
I'm trying to find out dy/dx of the function on the attachment using implicit differentiation, apparently some of my answers to this have different sign that the text book answer i've done it over and over again but still can't find the stage that goes wrong in my working. would you please have a look at my working on the attachment and let me know if I'm wrong at a certain stage

12789875_1989066897984304_1415500487_o.jpg137.2KB

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Reply 1

Zacken

Original post by Alen.m

1. This would be easier done if you factorised out

\cos y

and used the product rule:

\cos y(\cos x + \sin x) = \frac{1}{2}

2. I'm not sure where your first term is coming from? What have you differentiated to get

\cos x

Reply 2

MathsFanatic!

Hi, I cant open the image...
what exam board are you doing?

Reply 3

Zacken

Original post by Alen.m

...

Given that I don't think what you're doing is currently right, I'll lead you down the right path:

\cos y(\cos x+ \sin x) = \frac{1}{2}

.

Differentiate both sides w.r.t x:

\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}( \cos y (\cos x + \sin x)) = 0

Product rule on the LHS:

\displaystyle (\cos x + \sin x)\frac{\mathrm{d}}{\mathrm{d}x}(\cos y) + \cos y\frac{\mathrm{d}}{\mathrm{d}x}(\cos x + \sin x) = 0

Reply 4

Alen.m

Original post by Zacken

1. This would be easier done if you factorised out

\cos y

and used the product rule:

\cos y(\cos x + \sin x) = \frac{1}{2}

2. I'm not sure where your first term is coming from? What have you differentiated to get

\cos x

i actually used product rule for the d/dx cosy cos x as i've shown in the attachment

12789792_1989107224646938_714209742_o.jpg169.2KB

Reply 5

Alen.m

Original post by Student1408

Hi, I cant open the image...
what exam board are you doing?

AQA page 86 question 10 part b

Reply 6

Zacken

Original post by Alen.m

i actually used product rule for the d/dx cosy cos x as i've shown in the attachment

\cos y - \sin y \frac{dy}{dx}

does not mean cos y multiplied by sin y multiplied by dy/dx. You'll need to include brackets to make that a multiplication. As it stand, that's a subtraction. :tongue:

Reply 7

Alen.m

Original post by Zacken

\cos y - \sin y \frac{dy}{dx}

does not mean cos y multiplied by sin y multiplied by dy/dx. You'll need to include brackets to make that a multiplication. As it stand, that's a subtraction. :tongue:

can't find cos y- sin y which part of it you mean?

Reply 8

B_9710

Original post by Alen.m

I said this sometime last week, you do not find dy/dx of a function.
Here you could find dy/dx by differentiating with respect to x and then rearranging the find dy/dx. But you do not find dy/dx of a function - it doesn't make sense.

Reply 9

Alen.m

Original post by B_9710

im not trying to find dy/dx here . I'm trying to differentiate cos y cos x+ cos y sin x =1/2 using product rule and then rearrange to find dy/dx

Reply 10

B_9710

Original post by Alen.m

im not trying to find dy/dx here . I'm trying to differentiate cos y cos x+ cos y sin x =1/2 using product rule and then rearrange to find dy/dx

To find dy/dx, so you are trying to find dy/dx.

\displaystyle \frac{d}{dx} (\cos y \cos x)=\sin y \sin x \frac{dy}{dx}

Reply 11

Alen.m

Original post by B_9710

To find dy/dx, so you are trying to find dy/dx.

\displaystyle \frac{d}{dx} (\cos y \cos x)=\sin y \sin x \frac{dy}{dx}

did you use product rule to find that?i use product rule but i keep getting wrong answer

Reply 12

Alen.m

Original post by B_9710

To find dy/dx, so you are trying to find dy/dx.

\displaystyle \frac{d}{dx} (\cos y \cos x)=\sin y \sin x \frac{dy}{dx}

This is how i done it

image.jpg468.9KB

Reply 13

B_9710

Original post by Alen.m

did you use product rule to find that?i use product rule but i keep getting wrong answer

My mistake

\displaystyle \frac{d}{dx}(\cos y\cos x)=-(\cos x\sin y) \left (\frac{dy}{dx} \right ) -\sin x\cos y

.
I forgot to add the last bit.

(edited 8 years ago)

Reply 14

Alen.m

Original post by B_9710

My mistake

\displaystyle \frac{d}{dx}(\cos y\cos x)=-(\cos x\sin y) \left (\frac{dy}{dx} \right ) -\sin x\cos y

.
I forgot to add the last bit.

I've managed to diffrentatiate the whole equation but when it comes to rearenging to find dy/dx i just cant get the same answer as what the text book says . I've attached the text book answer and also the whole diffrentation that needs to be rearenged but i just cant get the write answer

Reply 15

Alen.m

Original post by Zacken

\cos y - \sin y \frac{dy}{dx}

does not mean cos y multiplied by sin y multiplied by dy/dx. You'll need to include brackets to make that a multiplication. As it stand, that's a subtraction. :tongue:

Attachment not found

image.jpeg389.1KB

Reply 16

Excuse Me!

Original post by Alen.m

Attachment not found

When you've completed the differentiation than you need to move everything that does not have DY/DX to the right hand side.

Factor out DY/DX and then divide the right hand side by whats in the brackets.

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