How was the chain rule used to find the derivative of a trigonometric function?

In summary: So the final answer would be "In summary, the equation is being solved using the chain rule for derivatives and the final answer is 3L."
  • #1
PrudensOptimus
641
0
I didn't understand how they arrived at this step:

lim as x -> 0 { 3 cos 3x (dy/dx) - 3 sin 3x (dy/dx)}

from

lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }

can some one explain pls?
 
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  • #2
I doubt that anyone can explain how one step follows from the other because you haven't shown the whole steps.

I assume what you have says that "lim as x -> 0 { 3 cos 3x (dy/dx) - 3 sin 3x (dy/dx)}" EQUALS something and that that follows from the fact that "lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }" equals something else. Unfortunately, you haven't told us what they equal.

Since the only difference between the two that I can see is that the first above is multiplied by 3, I would suspect that

"lim as x -> 0 { 3cos 3x (dy/dx) - 3sin3x (dy/dx) }= 3L"

follows from
"lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }= L" by multiplying the equation by 3. But of course, I can't be sure.
 
  • #3
they equal to -3 sin3x.

I was just wondering how did they get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x...
 
  • #4
Please tell us exactly what it is that you (or "they") are trying to prove -- not just this one step, but the overall context.

And, what is the entire expression or equation you are starting out with?
 
  • #5
No, they DON'T "equal to -3 sin3x." They can't because each has a
"lim as x->0". Once again, I can't say precisely what happened because you still haven't given us the whole thing but the obvious way to "get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x" is to multiply by 3!
 
  • #6
PrudensOptimus said:
they equal to -3 sin3x.

I was just wondering how did they get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x...

To get that by getting the derivative of 3x which is 3 then copy cos3x and you will get
"3 cos 3x" same as "sin 3x to be 3 sin 3x". Not equal to -3 sin3x.
 
Last edited:
  • #7
if i understand what you mean, they just used the chain rule for derivatives.
 

What are the basic derivative rules for trigonometric functions?

The basic derivative rules for trigonometric functions are as follows:
1. The derivative of sine is cosine: d/dx(sin(x)) = cos(x)
2. The derivative of cosine is negative sine: d/dx(cos(x)) = -sin(x)
3. The derivative of tangent is secant squared: d/dx(tan(x)) = sec^2(x)
4. The derivative of cotangent is negative cosecant squared: d/dx(cot(x)) = -csc^2(x)
5. The derivative of secant is secant times tangent: d/dx(sec(x)) = sec(x)tan(x)
6. The derivative of cosecant is negative cosecant times cotangent: d/dx(csc(x)) = -csc(x)cot(x)

How do you find the derivatives of inverse trigonometric functions?

To find the derivatives of inverse trigonometric functions, you can use the following formula: d/dx(arcsin(u)) = 1 / sqrt(1-u^2) * du/dx
Where u is the inner function and du/dx is the derivative of the inner function. This formula can be applied to all inverse trigonometric functions, such as arccos, arctan, and arccot.

What is the chain rule for derivatives of trigonometric functions?

The chain rule for derivatives of trigonometric functions states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function. In other words, if f(x) and g(x) are functions, then the derivative of f(g(x)) is equal to f'(g(x)) * g'(x).

Can you find the derivatives of trigonometric functions using the quotient rule?

Yes, you can find the derivatives of trigonometric functions using the quotient rule. This rule states that the derivative of a quotient of two functions is equal to the denominator multiplied by the derivative of the numerator, minus the numerator multiplied by the derivative of the denominator, all over the square of the denominator. This rule can be used to find the derivatives of functions such as sin(x)/x, cos(x)/x, and tan(x)/x.

How do you apply the derivatives of trigonometric functions to real-world problems?

The derivatives of trigonometric functions can be applied to real-world problems in various fields such as physics, engineering, and finance. For example, in physics, the derivatives of trigonometric functions can be used to calculate the velocity and acceleration of a moving object. In engineering, they can be used to solve problems involving motion and forces. In finance, they can be used to analyze financial data and make predictions. Additionally, the derivatives of trigonometric functions are also important in the study of calculus, which is essential for many scientific and mathematical fields.

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