# Trigonometry differentiation problem

• bryansonex
In summary, the first formula says that 2 sin(x) is what you're trying to solve for, but when you take the derivative, you end up with 1 - sin(x). The second formula says that if you want to solve for sin(x) you need to take the derivative of cos(x), and when you do this you get -2 sin(x)
bryansonex

## Homework Statement

Differentiate the following
$$cos^2 x$$

## The Attempt at a Solution

$$\frac{d(cos^2 x)}{dx} = 2 (-sin^{2-1} x).\frac{dx}{dx} = 2 (-sin x).1 = -2 sin x$$

$$-2 cos x sin x$$

Can you please tell me what's wrong !

You should be more careful in applying the chain rule. There are many formulations, the way I prefer it is to let u = cos(x). Then you are asked for d(u2(x))/dx. According to the chain rule, this is
$$\frac{d}{dx} u(x)^2 = \frac{d(u^2)}{du} \cdot \frac{du}{dx}$$
The first factor is 2u = 2 sin(x), but then you need to multiply du/dx, not dx/dx.

Or another rule of thumb which I prefer is to always take the derivative of what's inside... I'll show you what I mean:

If we are to take the derivative of $x^n$ we get $nx^{n-1}$
but behind the scenes you're also taking the derivative of x, but since this is 1, multiplying by 1 doesn't change anything.

Now if we let x be anything else, such as sinx then we have $(sinx)^n$ and taking the derivative of this is the same as before, but this time the derivative of what's inside (in this case, sinx) is cosx so this makes a difference: $n(sinx)^{n-1}.cosx$

But you also need to dig deeper and keep taking the derivative of what's inside, if there is something further inside.

e.g. $\left(sin(1/x)\right)^n$
Again, take the derivative by first applying the power rule, and then taking the derivative of sin(1/x) by using the trigo change, so cos(1/x) multiplies on the outside, but finally you also need to take the derivative of 1/x (this was also done behind the scenes in the 2nd example, we took the derivative of x, but again that's 1) so altogether we get:

$$n.\left(sin(1/x)\right)^{n-1}.cos(1/x).\frac{-1}{x^2}$$

And you can of course simplify. Anyway, I know this is terribly set out but I hope you understand it

## 1. What is the purpose of using trigonometry in differentiation?

Trigonometry is used in differentiation to calculate the rate of change of a function with respect to its independent variable. It allows us to find the slope of a curve at any point, which is essential in solving real-world problems involving motion, forces, and other physical phenomena.

## 2. What is the basic formula for differentiating trigonometric functions?

The basic formula for differentiating trigonometric functions is: d/dx(sin(x)) = cos(x) and d/dx(cos(x)) = -sin(x). This means that the derivative of sine is cosine, and the derivative of cosine is negative sine.

## 3. How do you handle other trigonometric functions such as tangent, cotangent, secant, and cosecant?

The derivatives of tangent, cotangent, secant, and cosecant can be found using the quotient rule and the chain rule. For example, the derivative of tangent is equal to the derivative of sine divided by the square of cosine.

## 4. Can you provide an example of solving a trigonometry differentiation problem?

Sure, let's say we have the function y = 2sin(x) + 3cos(x). To find its derivative, we use the basic formula for differentiating trigonometric functions. The derivative of 2sin(x) is 2cos(x) and the derivative of 3cos(x) is -3sin(x). Therefore, the derivative of y is dy/dx = 2cos(x) - 3sin(x).

## 5. What are some common mistakes to avoid when solving trigonometry differentiation problems?

One common mistake is forgetting to use the chain rule when differentiating trigonometric functions with a variable inside the function. Another mistake is not simplifying the derivative properly, leading to incorrect solutions. It is also important to pay attention to the signs of the trigonometric functions when using the quotient rule. Practice and careful attention to detail can help avoid these mistakes.

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