Jaweria
  • Jaweria
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Curry
  • Curry
just post, i got you.
Jaweria
  • Jaweria
ok first one is: (2x-3)^2+(4y-5)^2=10
Jaweria
  • Jaweria
and we need to find dy/dx for all of them.

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More answers

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

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