Finding Dy/Dx for expression

[RyanGall93]

 

[MarkFL]

I would likely begin by expression the equation as:

[MATH]\cos(y)=\sqrt{x}[/MATH]
And then implicitly differentiate. Can you proceed?

 

[RyanGall93]

I would likely begin by expression the equation as:

[MATH]\cos(y)=\sqrt{x}[/MATH]
And then implicitly differentiate. Can you proceed?

This is what’s i got for this question, not sure if correct or not?

 

[pka]

Another way is \(\displaystyle \frac{d}{du}\left(\arccos(u)\right)=-\frac{u'}{\sqrt{1-u^2}}\)

 

[MarkFL]

If I were to proceed using my suggestion, I would get:

[MATH]-\sin(y)\d{y}{x}=\frac{1}{2\sqrt{x}}[/MATH]
[MATH]\d{y}{x}=-\frac{1}{2\sqrt{x}\sin(y)}=-\frac{1}{2\sqrt{x-x^2}}[/MATH]
Do you see that you'd get the same from pka's suggestion?

Your method was correct, you just made some slips in the details. can you spot them?

 

[Steven G]

What does du/dx = ??

 

[RyanGall93]

If I were to proceed using my suggestion, I would get:

[MATH]-\sin(y)\d{y}{x}=\frac{1}{2\sqrt{x}}[/MATH]
[MATH]\d{y}{x}=-\frac{1}{2\sqrt{x}\sin(y)}=-\frac{1}{2\sqrt{x-x^2}}[/MATH]
Do you see that you'd get the same from pka's suggestion?

Your method was correct, you just made some slips in the details. can you spot them?

Nah I can’t. I don’t have a great eye for calculus I’m afraid.

 

[RyanGall93]

What does du/dx = ??

x^-1/2

 

[Steven G]

x^-1/2

That is not true. Consider what the derivative of x³ is wrt x, is it x²?