Trig Derivative

[Jason76]

Post Edited

\(\displaystyle y = \cot^{4} 2x\)

\(\displaystyle y' = 4 (\cot^{3} 2x) (- \csc^{2} 2x) 2\) :-? The cotangent trig name isn't changed on the left, but changes in other places.

\(\displaystyle y' = - 8 \cot^{3} 2x \csc^{2} 2x\)

Ok, I see the logic behind this, but note this is different than:

\(\displaystyle y = \cot 2x\)

\(\displaystyle y' = -\csc^{2} 2x (2)\) The cotangent trig name is changed (to cosecent). There are no instances of it staying the same anywhere.

\(\displaystyle y' = -2 \csc^{2} 2x\)

Ok, I see how this is done (when you see a trig function to a power other than 1, you don't change the trig function name) and can replicate. But don't really understand the logic behind the trig function name changing.

 

[srmichael]

\(\displaystyle y = \cot^{4} 2x\)

\(\displaystyle y' = 4 (\cot^{3} 2x) (- \csc^{2} 2x) 2\) :-? The cotangent trig name isn't changed on the left, but changes in other places.

\(\displaystyle y' = - 8 \cot^{3} 2x \csc^{2} 2x\)

Ok, I see the logic behind this, but note this is different than:

\(\displaystyle y = \cot 2x\)

\(\displaystyle y' = \csc 2x (2)\) The cotangent trig name is changed (to cosecent). There are no instances of it staying the same anywhere. ====> INCORRECT!

\(\displaystyle y' = 2 \csc 2x\) ====> INCORRECT!

Ok, I see how this is done (when you see a trig function to a power other than 1, you don't change the trig function name) and can replicate. But don't really understand the logic behind the trig function name changing.

Cotangent is changing to cosecant when you take the derivative of cotangent because that is the formula.

If y = cot(x) then y' = -csc²(x)

With that formula, make sure you know these as well:

If y = tan(x) then y' = sec²(x)

If y = sin(x) then y' = cos(x)

If y = cos(x) then y' = -sin(x)

If y = csc(x) then y' = -csc(x)cot(x)

If y = sec(x) then y' = sec(x)tan(x)

 

[Jason76]

Cotangent is changing to cosecant when you take the derivative of cotangent becuase that is the formula.

If y = cot(x) then y' = -csc²(x)

With that formula, make sure you know these as well:

If y = tan(x) then y' = sec²(x)

If y = sin(x) then y' = cos(x)

If y = cos(x) then y' = -sin(x)

If y = csc(x) then y' = -csc(x)cot(x)

If y = sec(x) then y' = sec(x)tan(x)

Original Post Edited.

 

[srmichael]

You're missing the underlying concept. When you take the derivative of say y = cot²(x), you are using a combination of the Power Rule and the Chain Rule.

Power Rule: \(\displaystyle y=x^n \ then\ y'=n\cdot x^{n-1}\)

Chain Rule: \(\displaystyle y=f(g(x)) \ then\ y'=f'(g(x))\cdot g'(x)\)

So for cot²(x), f(x) is x² and g(x) is cot(x)

 

[DrPhil]

Post Edited

\(\displaystyle y = \cot^{4} 2x\)

\(\displaystyle y' = 4 (\cot^{3} 2x) (- \csc^{2} 2x) 2\) :-? The cotangent trig name isn't changed on the left, but changes in other places.

\(\displaystyle y' = - 8 \cot^{3} 2x \csc^{2} 2x\)

Ok, I see the logic behind this, but note this is different than:

\(\displaystyle y = \cot 2x\)

\(\displaystyle y' = -\csc^{2} 2x (2)\) The cotangent trig name is changed (to cosecent). There are no instances of it staying the same anywhere.

\(\displaystyle y' = -2 \csc^{2} 2x\)

Ok, I see how this is done (when you see a trig function to a power other than 1, you don't change the trig function name) and can replicate. But don't really understand the logic behind the trig function name changing.

The idea of "name changing" is not useful. The only function that does NOT "change its name" when differentiated is the exponential:

\(\displaystyle \displaystyle \frac{d}{dx}\left( e^x\right) = e^x\)

No other function besides the exponential is its own derivative.

The reason there is still a cotangent in the first example is because of the power law.

\(\displaystyle \displaystyle \frac{d}{dx}\left( u^n\right) = n\ u^{n-1}\ \frac{du}{dx}\)

The "logic" of trigonometric derivatives is that the derivative of the sine is the cosine, and the the derivative of the cosine is the negative of the sine. All other derivatives of trig functions follow from those two - which are the only ones I have memorized. There is symmetry and beauty in that truth. That's all I know on earth, and all I need to know.

 

[Jason76]

You're missing the underlying concept. When you take the derivative of say y = cot²(x), you are using a combination of the Power Rule and the Chain Rule.

Power Rule: \(\displaystyle y=x^n \ then\ y'=n\cdot x^{n-1}\)

Chain Rule: \(\displaystyle y=f(g(x)) \ then\ y'=f'(g(x))\cdot g'(x)\)

So for cot²(x), f(x) is x² and g(x) is cot(x)

But when doing the power rule on \(\displaystyle \cot^{4} x\), it does not become \(\displaystyle 4(-\csc^{3} x)\)

 

[HallsofIvy]

But when doing the power rule on \(\displaystyle \cot^{4} x\), it does not become \(\displaystyle 4(-\csc^{3} x)\)

NO! It doesn't! That's the whole point of the chain rule. The derivative of \(\displaystyle \cot^4(x)\) is \(\displaystyle 4 cot^3(x)\) times the derivative of cot(x) which is -csc(x).

 

[srmichael]

NO! It doesn't! That's the whole point of the chain rule. The derivative of \(\displaystyle \cot^4(x)\) is \(\displaystyle 4 cot^3(x)\) times the derivative of cot(x) which is -csc(x).

I think HOI meant to say was the derivative of cot(x) is -csc²(x).

 

[Deleted member 4993]

But when doing the power rule on \(\displaystyle \cot^{4} x\), it does not become \(\displaystyle 4(-\csc^{3} x)\)

Because, the rule of differentiation is:

if

y = [f(x)]ⁿ

we have

y' = (n) * [f(x)]^n-1 * f'(x)

here

f(x) = cot(x)

similar rule will apply for any proper function. For example,

y = [ln(x)]⁷

y' = 7 * [ln(x)]⁶ * (1/x)