# Thread: Implicit Differentiation Problem (Chain Rule?)

1. ## Implicit Differentiation Problem (Chain Rule?)

Find dy/dx:
y=[sin(x+5)]^(5/4)

So far I have this:
y=[sin(x+5)]^(5/4)
y'={[(5/4)cosx(x+5)]^(1/2)} [cos(x+5)] (1)

There are only answers to odd numbered questions in the back of my textbook & this is an even numbered problem. I'm not sure where to go from here or if I've used chain rule the right way.

2. Couple of questions:
1. Why did the first sin() magically turn into cos()?
2. Why did fourths magically turn into halves?
3. What about this is "implicit"?

Please give it another go.

3. Originally Posted by veronicadeno
Find dy/dx:
y=[sin(x+5)]^(5/4)

So far I have this:
y=[sin(x+5)]^(5/4)
y'={[(5/4)cosx(x+5)]^(1/2)} [cos(x+5)] (1)

There are only answers to odd numbered questions in the back of my textbook & this is an even numbered problem. I'm not sure where to go from here or if I've used chain rule the right way.
Except for the peculiarities that tkhunny asked about, it looks like you did the chain-rule part ok! It is just the power-law that is a "bit" off.

4. Originally Posted by tkhunny
Couple of questions:
1. Why did the first sin() magically turn into cos()?
2. Why did fourths magically turn into halves?
3. What about this is "implicit"?

Please give it another go.
1. Because I thought the derivative of sin() was cos(), unless I'm not supposed to mess with the trig in that first part which is what I think you're implying.
2. I think that is a typing error on my part, should be 1/4 I think, but then again I'm in a math help forum.
3. No idea, doesn't look implicit to me, that was just the heading in my textbook.
Thank you!

Originally Posted by DrPhil
Except for the peculiarities that tkhunny asked about, it looks like you did the chain-rule part ok! It is just the power-law that is a "bit" off.
Okay thank you!

"Another go"
y=[sin(x+5)]^(5/4)
y'={[(5/4)cosx(x+5)]^(1/4)} [cos(x+5)] (1)
y'=
{[(5/4)cosx(x+5)]^(1/4)} [cos(x+5)]

5. Originally Posted by veronicadeno
1. Because I thought the derivative of sin() was cos(), unless I'm not supposed to mess with the trig in that first part which is what I think you're implying.
2. I think that is a typing error on my part, should be 1/4 I think, but then again I'm in a math help forum.
3. No idea, doesn't look implicit to me, that was just the heading in my textbook.
Thank you!

Okay thank you!

"Another go"
y=[sin(x+5)]^(5/4)
y'={[(5/4) sin(x+5)]^(1/4)} [cos(x+5)] (1)
y'=
{[(5/4) sin(x+5)]^(1/4)} [cos(x+5)]
Good answers.

1. Yes, the first part is just the power-law - no trig - so still should be sin(x+5)
2. Yes, the power is 5/4 - 1 = 1/4
3. Yes, this function is explicit, not implicit

EDIT: Thank you srmichael for fixing my blunder!

y'= {[(5/4) sin(x+5)]^(1/4)} [cos(x+5)]

6. Originally Posted by DrPhil
Good answers.

1. Yes, the first part is just the power-law - no trig.
2. Yes, the power is 5/4 - 1 = 1/4
3. Yes, this function is explicit, not implicit

Only remaining question is what you mean by cosx(x+5). That extra letter does NOT belong. The answer is

y'= {[(5/4)cos(x+5)]^(1/4)} [cos(x+5)]
There's a liittle hiccup in Dr. Phil's answer:

$y=[\sin(x+5)]^{\frac{5}{4}}$

$y'=\frac{5}{4}[\sin(x+5)]^{\frac{1}{4}}\cos(x+5)$

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