Implicit Differentiation: differentiate y = sin(4x + 5y), then find dy/dx

[Vulcan]

The problem is to differentiate y = sin(4x + 5y) then find dy/dx!

d/dx sin(4x + 5y) = 4cos(4x + 5y) not sure where to go next!

 

[Deleted member 4993]

The problem is to differentiate y = sin(4x + 5y) then find dy/dx!

d/dx sin(4x + 5y) = cos(4x + 5y) * [4 + 5(dy/dx)]

not sure where to go next!

continue.....

 

[stapel]

The problem is to differentiate y = sin(4x + 5y) then find dy/dx!

d/dx sin(4x + 5y) = 4cos(4x + 5y) not sure where to go next!

I'm not sure what you're doing here...?

Instead, start with the original equation:

. . . . .$\displaystyle y\, =\, \sin(4x\, +\, 5y)$

Then differentiate both sides, remembering that dx/dx = 1 (so we usually ignore it, as we will here) but dy/dx remains dy/dx:

. . . . .$\displaystyle \dfrac{dy}{dx}\, =\, \cos(4x\, +\, 5y)\, \cdot\, \left(4\, +\, 5\, \dfrac{dy}{dx}\right)$

Then solve for "dy/dx =".

 

[Vulcan]

Yeah, after I posted I came up with 1 = cos(4x+5y)(4+5y dy/dx) then 1 = 4cos(4x+5y)(5 dy/dx) but not sure how to get to the solution which is

4cos(4x+5y)/1-5cos(4x+5y)

Incidentally how do you get the proper format in this window, when I try it the dy/dx disappears

Thanks

 

[stapel]

Yeah, after I posted I came up with

1 = cos(4x+5y)(4+5y dy/dx)

What happened to the derivative of the "y" (that is, the "dy/dx") on the left-hand side? Put that derivative back!

then

1 = 4cos(4x+5y)(5 dy/dx)

What happened to the "plus" between the "4" and the "5 dy/dx" in the second parenthetical in the first quote above? Put that sign back!

but not sure how to get to the solution

Use the help you're given. I gave you:

. . . . .$\displaystyle \dfrac{dy}{dx}\, =\, \cos(4x\, +\, 5y)\, \cdot\, \left(4\, +\, 5\, \dfrac{dy}{dx}\right)$

Did you see this? If so, how did you get your result from this? Or did you start over? What was your reasoning? What were your steps?

Next, use algebra. Starting from what I gave you:

. . . . .$\displaystyle \dfrac{dy}{dx}\, =\, \cos(4x\, +\, 5y)\, \cdot\, \left(4\, +\, 5\, \dfrac{dy}{dx}\right)$

...you know that you're wanting to solve for "dy/dx =". For the time being, let's rename "dy/dx" as "D". Then the equation becomes:

. . . . .$\displaystyle D\, =\, \cos(4x\, +\, 5y)\, \cdot\, \left(4\, +\, 5\,D\right)$

To get the target variable "D" by itself, let's do that multiplying-out thing that you seemed to be trying:

. . . . .$\displaystyle D\, =\, 4\, \cos(4x\, +\, 5y)\, +\, 5\, \cos(4x\, +\, 5y)\, D$

Now apply algebraic techniques to gather together the terms containing "D", factor, and divide through. What do you get?

Please show your work. Thank you!

 

[Vulcan]

Stapel

I see where I went wrong, when I differentiated the y on the LHS it should have been dy/dx then I dropped the + sign. Why I don't know. After I sorted that the algebraic manipulation was straightforward. Just getting back into calculus after 30+ years so I am extremely rusty.

Many thanks for your help.