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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.

Calculus1
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just post, i got you.
ok first one is: (2x-3)^2+(4y-5)^2=10
and we need to find dy/dx for all of them.

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Other answers:

you can either use implicit differentiation from the start, or you can begin by isolating the y variable.
second question is: x^2sin(x)+y^2cos(y)=1
likewise for the second
yeah we have to do implicit diffentiation but I am really stuck in all of them.
i'll get you started with (2x-3)^2+(4y-5)^2=10
ohh thank you so much
|dw:1359266294450:dw|
idk if you can read the edge of it, it should say = 0
yeah i can
but after the paranthesis is that a multiply by 2 or its a power of 2 for the first one
multiplication
ok
but then we have to simplify more right?
yeah you would have to isolate y'
ok can we go through this together till the end?
sure
thank u :)
|dw:1359266644304:dw|
|dw:1359266753423:dw|
how did u get -2?
oh my bad, that should be a 12
|dw:1359266876031:dw|
oh ok
|dw:1359266992506:dw|
and do we have to put y^1 before paranthesis?
where?
its ok I got it
ok. try the second one whenever you are comfortable
ok
ok I have a feeling that I ll again get stuck with this one:
x^2sin(x)+y^2cos(y)=1
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
ok
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)]
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.
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 :(
I m still learning
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 ?
2x cos(x)
are you sure?
not much
don't forget the product rule
i always mess up on product rule and chain rule
\[\frac{ d }{ dx } x^{2} \sin x = x^{2} \cos x + \sin x * 2x\]
"first times the derivative of the second, plus second times the derivative of the first"
ohh ok
now try differentiating y^2cos(y)
ok
not sure if I did it right or not
|dw:1359268366069:dw|
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
oh ok
so i think you got that right, too, just needed the dy/dx ?
yes
\[x^{2} \cos x + 2x \sin x + \frac{ dy }{ dx } * -y^{2} \sin y + \frac{ dy }{ dx } * 2y \cos y\]
so that will be |dw:1359268690019:dw|
|dw:1359268763927:dw|
ok
you need to isolate dy/dx now
yeah ok
|dw:1359269043320:dw|
is that right?
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
ok thanks
wait let me double check something
couple more please if you dont mind?
ok
ok. sorry. if you had more questions you might want to start a new post
oh ok
after a few of these, implicit differentiation will be easy for you
this is the first time ever I am using this site can you suggest me that how to start new post?
Yeah I am trying my best.

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