Derivative for sin(x)/1+cos(x): Why is it 1/1+cos(x) but not -1/1+cos(x)?

[coooool222]

Why is it 1/1+cos(x) but not -1/1+cos(x)

 

[skeeter]

[imath]\dfrac{d}{dx} \bigg[\dfrac{\sin{x}}{1+\cos{x}}\bigg] =[/imath]

[imath]\dfrac{(1+\cos{x})\cos{x} - \sin{x}(-\sin{x})}{(1+\cos{x})^2}[/imath]

try again ...

 

[coooool222]

[imath]\dfrac{d}{dx} \bigg[\dfrac{\sin{x}}{1+\cos{x}}\bigg] =[/imath]

[imath]\dfrac{(1+\cos{x})\cos{x} - \sin{x}(-\sin{x})}{(1+\cos{x})^2}[/imath]

try again ...

why cant i take the derivative of 1+cos(x) first?

 

[skeeter]

why cant i take the derivative of 1+cos(x) first?

because that's not the quotient rule ...

[math]\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}[/math]

 

[coooool222]

because that's not the quotient rule ...

[math]\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}[/math]

is it just me or does this equation always change. sometimes i see g'(x) first and sometimes f'(x) first.

 

[skeeter]

it hasn't changed in the 20+ years I've taught the course ...

[math]\left(\dfrac{\text{top}}{\text{bottom}}\right)' = \dfrac{\text{bottom times derivative of the top} - \text{top times derivative of the bottom}}{(\text{bottom})^2}[/math]

 

[Dr.Peterson]

is it just me or does this equation always change. sometimes i see g'(x) first and sometimes f'(x) first.

The product rule can be written in any order, because it's all addition and multiplication; many books will put g' first there.

Because the quotient rule involves subtraction, order matters; f' must be in the first term. You may be confusing the latter with the former.

That's why I teach writing both in the same order, to make it harder to get the quotient rule wrong.