Question

find dy/dx by implicit differentiation: siny=x^2-e^y

Answer 1

\[\frac{d}{dx}(\sin(f(x))=?\]

\[\frac{d}{dx}(x^2)=?\]
\[\frac{d}{dx}e^{f(x)}=?\]

Answer 2

The first and last problem I just listed, you will need to know chain rule.
The middle one is just applying the power rule.
The first one along with chain rule you will need to know the derivative of sin(x) is cos(x).
The last one along with chain rule you will need to know the derivative of e^x is e^x.

(f(x))'=f'(x) or df(x)/dx or df/dx or dy/dx

Answer 3

Here are some of examples of chain rule:
---
\[\frac{d}{dx}\cos(x^2+1)=(x^2+1)'(-\sin(x^2+1))=(2x+0)(-\sin(x^2+1))=\]
\[=(2x)(-\sin(x^2+1))=-2xsin(x^2+1) \]

---

\[\frac{d}{dx}\sin(x^6)=(x^6)'\cos(x^6)=6x^5 \cos(x^6)\]

---

\[\frac{d}{dx} (f \circ g)(x)=\frac{d}{dx}f(g(x))=g'(x) \cdot f'(g(x)) \text{ or } =g'(x)(f' \circ g)(x)\]

Answer 4

Do you think you can attempt answering the 3 problems I stated in my fist post?

Answer 5

my main question is does e^y turn into e^x when using implicit differentiation? or does it change to e^y*y' and i have to solve from there?

Answer 6

The derivative of y is y'.
(e^y)'=y' * e^y

Answer 7

or you can write it as e^y * y' like you have done so.

Answer 8

Ok, so far what I have for the equation is:

siny=x^2-e^y

cosy*y'=2x-e^y*y'

I'm stuck in solving the rest of the problem because of the e^y*y'. I don't know what the next step would be.

Answer 9

Put your terms that have y' on one side of the equation and the other terms that don't contain y' on the on the side.
Just like if we have 5x=5-8x
We get our like terms together:
5x+8x=5

Factor x out like this:
x(5+8)=5

Then divide both sides by 5+8
So we get x=5/(5+8)

Answer 10

Ok I have:

cosy*y'+e^y*y'=2x

Now how do I get the e^y*y' as just y' so I can divide the cosy to get y' by itself.

Answer 11

Did you look at my example any?
If you are able to follow my example, you will see what to do.

Answer 12

Like if not, like tell me what you don't get.
Like I have 5x+8x=5
I want to solve for x.
I factor the x out like:
x(5+8)=5

Then to undo the multiplication of (5+8) by dividing both sides by (5+8).

Answer 13

So the result will be:

y'=2x/cosy+e^y

Answer 14

y'=2x/(cosy+e^y) <---is correct.

Answer 15

Ok, thank you so much for your help.

Answer 16

Np. Do you understand what I was doing with my linear equation? Do you see how knowing how to solve a linear equation applies here?

Answer 17

yea, I understood how you did all of that perfectly. Was just confused because e is raised to the y times y' power. I thought I would have to do something else to eliminate y' as part of the power for e.

Answer 18

Oh no y' is not part of the exponent
I thought you said (e^y)'=y'*e^y

Answer 19

not e^(y*y') <--this is a big no no)

Answer 20

\[\frac{d}{dx}e^{u(x)}=u'(x) \cdot e^{u(x)}\]

Answer 21

oh ok. Thank you, the exponential stuff always confuses me.

Answer 22

Well I hope it is less confusing now.

Answer 23

i'm slowly getting it

Answer 24

Let me know what questions you have.

Answer 25

will do, thank you

Answer 26

ok can you verify if I made a mistake somewhere:

question is use implicit differentiation to find dy/dx:
x^3+xy=y^2+x

3x^2+[x(y*y')+(1)(y)]=2y*y'+(1)
3x^2+xy*y'+y=2y*y'+1
3x^2+y-1=2y*y'-xy*y'
3x^2=y-1/2y-xy=(2y-xy)y'/2y-xy

y'=3x2=y-1/2y-xy

Answer 27

(xy)'=(x)'y+x(y)'
=1 y+x y'

Answer 28

oops i meant 3x^2+y-1/2y-xy

Answer 29

is the solution i have down correct? (the second one i sent) or do i need to fix something?

Answer 30

Well I think you still didn't apply what I said.
You denominator is a little off because of it.
And all I think you mean (3x^2+y-1)/(2y-xy) right?

Answer 31

Yes that's what I meant.

Answer 32

Did you look at what I said about the product rule?
(xy)'=(x)' * y + x * (y)'

Remember derivative or y is just y' not y*y'

Just like if I said find y' for y=x+1, you would say y'=1 not y*y'=1
or y'=1/y

The derivative of y is just y'

(xy)' = (x)' * y + x * (y)'
= 1 * y + x * y'

Answer 33

oh ok, so fixing it i got:

y'=(3x^2+y-1)/(2y-x)

Answer 34

Beautiful.

Answer 35

thank you. and I misread what you had typed, sorry.

thank you so much for your help.

Answer 36

About the product rule?

Answer 37

yes

Answer 38

Ok but after re-reading it, you understand the product rule and how you should apply if you have a product of functions.
The only time you can ignore the product rule is if one of those functions or both are constant functions but yeah you could still use the product rule then.
(xy)'=(x)' * y + x * (y)'


....


assume c is constant like you know 5 and sqrt(3) and pi (these are all constants)
5 will always be 5. it doesn't change. that is why it is called constant, it is constantly 5 all the time.

(cx)' = (c)' * x + c * (x)'
= 0 *x + c* (x)'
= c(x)'

or just use constant multiple rule (cx)'=c(x)'

Answer 39

Ok I have another one.

find the derivative of the function:
f(x)=4arcsin(x/2)

I have so far:
f'(x)=4[(1/2)/(squareroot 1-(x^2/4))

I don't know if I have this correct and stuck on what to do from here. Would by multiply the 4 and 1/2 to get:

2/(squareroot 1-(x^2/4))?

Answer 40

\[f'(x)=4 \cdot \frac{\frac{1}{2}}{\sqrt{(1-(\frac{x}{2})^2})}=4 \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{(1-\frac{x^2}{4}})}\]
\[=2 \cdot \frac{1}{\sqrt{(\frac{4}{4}-\frac{x^2}{4})}}=\frac{2}{\sqrt{(\frac{4-x^2}{4})}}\]
This should help.
That is square root of[ (4-x^2)/4]

For some reason the square root isn't displaying as I like.

Answer 41

I didn't do the whole thing on simplifying. I left some of it to you.
Let me know if you don't know what to do next.

Answer 42

I don't know if this is right but I get \[\frac{ \sqrt{4+x^2} }{4-x^2 }\]

Answer 43

How did you get that + sign?
Also I think a multiple of 4 disappeared from your answer?

Answer 44

I took the conjugate of \[\sqrt{4-x^2}\] to cancel the radical.

Answer 45

\[\sqrt{4-x^2} \cdot \sqrt{4-x^2}=(4-x^2)=4-x^2\]
just like \[\sqrt{9} \cdot \sqrt{9}=9 ;\sqrt{9} \cdot \sqrt{-9} \cancel{=} 9\]
Also are you required to rationalize the bottom?

Answer 46

ok as far as i can get then is

2/square root 4-x^2/2

Answer 47

so would the answer be

\[\frac{ 4\sqrt{4-x^2} }{ 4-x^2 }\]