I need help with 11 questions of Derivatives problems in Calculus 1. Can anyone help me with that please let me know, I will really appreciate it.

- Jaweria

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- Curry

just post, i got you.

- Jaweria

ok first one is: (2x-3)^2+(4y-5)^2=10

- Jaweria

and we need to find dy/dx for all of them.

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## More answers

- anonymous

you can either use implicit differentiation from the start, or you can begin by isolating the y variable.

- Jaweria

second question is: x^2sin(x)+y^2cos(y)=1

- anonymous

likewise for the second

- Jaweria

yeah we have to do implicit diffentiation but I am really stuck in all of them.

- anonymous

i'll get you started with (2x-3)^2+(4y-5)^2=10

- Jaweria

ohh thank you so much

- anonymous

|dw:1359266294450:dw|

- anonymous

idk if you can read the edge of it, it should say = 0

- Jaweria

yeah i can

- Jaweria

but after the paranthesis is that a multiply by 2 or its a power of 2 for the first one

- anonymous

multiplication

- Jaweria

ok

- Jaweria

but then we have to simplify more right?

- anonymous

yeah you would have to isolate y'

- Jaweria

ok can we go through this together till the end?

- anonymous

sure

- Jaweria

thank u :)

- anonymous

|dw:1359266644304:dw|

- anonymous

|dw:1359266753423:dw|

- Jaweria

how did u get -2?

- anonymous

oh my bad, that should be a 12

- anonymous

|dw:1359266876031:dw|

- Jaweria

oh ok

- anonymous

|dw:1359266992506:dw|

- Jaweria

and do we have to put y^1 before paranthesis?

- anonymous

where?

- Jaweria

its ok I got it

- anonymous

ok. try the second one whenever you are comfortable

- Jaweria

ok

- Jaweria

ok I have a feeling that I ll again get stuck with this one:

- Jaweria

x^2sin(x)+y^2cos(y)=1

- anonymous

just think of differentiating y like you were differentiating any other variable (except x). remember to write the "dy/dx" or " y' " after you differentiate

- Jaweria

ok

- anonymous

like here's an example:
if i differentiate y*sin y with respect to x:
= y cos(y) * y' + sin(y) * y'
= y' [y cos(y) + sin(y)]

- anonymous

it's the same as if you were differentiating 'u', as in
differentiate: sin(x^2)
let u = x^2 and differentiate sin(u)
d/dx sin(u) = cos(u) * du = cos(x^2) * du
Another way of saying that is: cos(x^2) * u'
The only difference here is, we can write du in terms of x very easily. With implicit differentiation, we normally can't do that, but it's the same concept.

- Jaweria

hmmm if you dont mind can we go over this problem because my brain is not working at all with all of these i am sorry but thats the truth all my 11 questions are due tomorrow online :(

- Jaweria

I m still learning

- anonymous

x^2sin(x)+y^2cos(y)=1
let's start with the easy part. if we're going to differentiate both sides, the right side becomes 0. what does x^2 sin(x) become ?

- Jaweria

2x cos(x)

- anonymous

are you sure?

- Jaweria

not much

- anonymous

don't forget the product rule

- Jaweria

i always mess up on product rule and chain rule

- anonymous

\[\frac{ d }{ dx } x^{2} \sin x = x^{2} \cos x + \sin x * 2x\]

- anonymous

"first times the derivative of the second, plus second times the derivative of the first"

- Jaweria

ohh ok

- anonymous

now try differentiating y^2cos(y)

- Jaweria

ok

- Jaweria

not sure if I did it right or not

- Jaweria

|dw:1359268366069:dw|

- anonymous

the left part is almost right, just need the negative sign (the derivative of cos is -sin
the other part would become:
cos(y) * 2y * dy/dx

- Jaweria

oh ok

- anonymous

so i think you got that right, too, just needed the dy/dx ?

- Jaweria

yes

- anonymous

\[x^{2} \cos x + 2x \sin x + \frac{ dy }{ dx } * -y^{2} \sin y + \frac{ dy }{ dx } * 2y \cos y\]

- Jaweria

so that will be |dw:1359268690019:dw|

- anonymous

|dw:1359268763927:dw|

- Jaweria

ok

- anonymous

you need to isolate dy/dx now

- Jaweria

yeah ok

- Jaweria

|dw:1359269043320:dw|

- Jaweria

is that right?

- anonymous

yep perfect
you can eliminate the minus sign on the outside if you wanted to by swapping the terms in the denominator to make it:
y^2 sin y - 2y cos y
but thats just a matter of preference

- Jaweria

ok thanks

- anonymous

wait let me double check something

- Jaweria

couple more please if you dont mind?

- Jaweria

ok

- anonymous

ok. sorry. if you had more questions you might want to start a new post

- Jaweria

oh ok

- anonymous

after a few of these, implicit differentiation will be easy for you

- Jaweria

this is the first time ever I am using this site can you suggest me that how to start new post?

- Jaweria

Yeah I am trying my best.

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