Question

Need help on calc plz help
Given curve defined by x^2-siny=y+4
Show that dy/dx = 2x/(1+cosy)

Answer 1

So i got this far:
d/dx(x^2)-d/dx(siny)=d/dx(y)+d/dx(4)

Answer 2

But in the end i got
2x-1/(cosy)=y^1

Answer 3

what did i do wrong?

Answer 4

The first step is this:
\[\huge \frac{dy}{dx}\left(x^2-\sin y\right)=\frac{dy}{dx}(y+4)\]
Then taking derivatives, assuming everywhere we see a \(y\) what we have is some (unknown) function of \(x\) that gives:
\[\huge 2x-\cos y \frac{dy}{dx}=\frac{dy}{dx}\]
Does that make sense so far?

Answer 5

wouldnt siny be a chain rule function?

Answer 6

When \[\huge \sin y\] is written what is actually there is \[\huge \sin f(x)\]
So when you take \[\huge \frac{dy}{dx}\sin y\] You're actually doing this:
\[\huge \frac{dy}{dx}\sin f(x) = \cos f(x) \cdot f'(x) = \cos f(x) \frac{dy}{dx}\]

Answer 7

Oh drat and fie. It ran off the edge of the screen! :(

Answer 8

so basically it would be cos(dy/dx) or as some write it as cos(y^1)?

Answer 9

OH POOP.
The derivative is
\[\huge -\sin(x^2-3x)\cdot(2x-3)\]

Answer 10

So \[\huge \frac{dy}{dx}\cos f(x) = -\sin f(x)\cdot f'(x)\]

Answer 11

This is the correct version of the example:
\[\huge \frac{dy}{dx}\cos (x^2-3x) = \cos f(x)\cdot (2x-3)\]

Answer 12

Refer to the Mathematica attachment.

Answer 13

Your attacment made sense, however the respnse given by mathteacher really threw me off. Confused me a lot.

Answer 14

But i dont understand how you got 1+ cosy on the bottom

Answer 15

Refer to the corrected Mathematica attachment.

Answer 16

@calchelpneededplz sorry for the confusion.
Basically you're using the chain rule.

when you take
\[\huge \frac{dy}{dx}\sin(y)\]
the result is \[\huge\cos(y)\cdot\frac{dy}{dx}\]
This is the chain rule. Lots of examples are given here: http://tutorial.math.lamar.edu/Classes/CalcI/ChainRule.aspx

In your problem you have to take dy/dx of both sides, then "solve for dy/dx" as if it were a variable. After correctly taking dy/dx on both sides and solving for dy/dx, the result is indeed 2x / ( 1 + cos y).

Answer 17

I know how to take the regular d/dx but im not quite familiar with taking dy/dx

Answer 18

Studying the chain rule will help.

dy/dx says "take the derivative of a function named "y" with respect to its variable, "x"."

When you have a function like \(y = \sin(x) \) This is easy, because dy/dx is just \(\cos(x)\).
But what if you had \(y = \sin (3x^2 - 4x)\) instead? What is dy/dx of that?

Answer 19

would it just be cos(3x^2-4x)(6x-4)

Answer 20

Right. Exactly. :)

But if we don't KNOW what function is "inside" sine... the general rule is

\[\frac{dy}{dx}\sin(y) = \cos(y) \cdot \frac{dy}{dx}\]

That says "the derivative of y with respect to x of sin of *some function y" is the derivative of sin(y) TIMES the derivative of whatever is inside sin(y).

You did this perfectly with cos(3x^2-4x)*(6x-4).

Answer 21

Check out the link I sent before about the chain rule. I gotta go now. Good luck! :)

Answer 22

okay thank you so much

Answer 23

@calchelpneededplz
"But i dont understand how you got 1+ cosy on the bottom"

Refer to the followup attachment.