Implicit Differentiation with Trig Functions

[JoeJ]

Could someone please show me step by step how to solve the following?

1. (sin (pi) * x + cos (pi) * y) ^ 2 = 2
2. sin x = x * (1 + tan y)
3. y = sin (x*y)

Thank you.

 

[soroban]

Hello, JoeJ!

Here's the first one.
. . That strange symbol (π) is "pi".

1. .[sin (πx) + cos(πy)]² .= .2
.

2[sin(πx) + cos(πx)] [π cos(πx) - π sin(πy)·y'] . = . 0

Then we have: . π cos(πx) - π sin(πy)·y' . = . 0

. . . - π sin(πy)·y' . = . - π cos(πx)

. . . . . . . . . . . . . . . . . cos(πx)
. . . . . . . . . . . y' . = . ----------
. . . . . . . . . . . . . . . . . sin(πy)

 

[JoeJ]

Ok, I got that. What about 2 & 3? For 2, I'm thinking...

(sin x) / x = (1 + tan y*y')
((sin x) / x) - 1 = tan y*y'
(((sin x / x) - 1) / (tan y) = y'

For 3...

y = sin (x*y)
y = sin (x) * sin (y*y')
(y / (sin (x)) = sin (y*y')

I know I'm doing something wrong here, but don't know. I'm thinking using the quotient rule for both?

 

[soroban]

Hello again, JoeJ!

2. sin x = x (1 + tan y)

Why would you want to introduce a quotient into the problem?
We have to use the Product Formula.

. . . . . . . cos x . = . x(sec²y)y') + 1 + tan y

. . . x(sec²y)y' . = . cos x - tan y - 1

. . . . . . . . . . . . . . . .cos x - tan y - 1
. . . . . . . . . .y' . = . ---------------------
. . . . . . . . . . . . . . . . . . x sec²y

3. y = sin (xy)

. . . . . . . . . . . . . . .y' . = . cos(xy) (xy' + y)

. . . . . . . . . . . . . . .y' . = . xy' cos(xy) + y cos(xy)

. . . . .y' - xy' cos(xy) . = . y cos(xy)

. . . y' [1 - x cos(xy)] . = . y cos(xy)

. . . . . . . . . . . . . . . . . . . . . .y cos(xy)
. . . . . . . . . . . . . . .y' . = . ----------------
. . . . . . . . . . . . . . . . . . . . 1 - x cos(xy)

 

[JoeJ]

Thank you very much. These helped me a lot on other problems similar to these.