# Finding the derivative of the function

• meeklobraca
In summary, the equation for homework is (cos5x)(5)(cosx^2) - (sin5x)(2Cosx) / cos x^3.The correct answer is (cos(x))^2.Thanks!
meeklobraca

## Homework Statement

f(x) = sin(5x) / cos(x^2)

## The Attempt at a Solution

I got 25Cosx-Sin 5xCosx / cos x^3

Would this be the correct answer?

Thanks!

Nope. Warm up with finding the derivatives of sin(5x) and cos(x^2). What are they? Now apply the quotient rule.

I have (cos5x)(5)(cosx^2) - (sin5x)(2Cosx) all divided by Cosx^4

Right track?

It's getting there. But what's the derivative of cos(x^2)? Chain rule. And cos(x^2)^2 is not cos(x^4).

Ahh I left out the -sin x

Okay so I should have

(cos5x)(5)(Cosx^2) - (sin5x)(2Cosx)(-sinx) / Cosx^4

?

Why don't you just do the pieces first? Derivative of cos(x^2)? You aren't going to get anywhere without that. Or did you actually mean to write cos(x)^2 in the denominator instead of cos(x^2)?

The derivative of cos x^2 is -2cosxsinx

Again, is that (cos(x))^2 or cos(x^2)? They are two different things. Correct if it's the first. Use parentheses!

Its the 2nd one. And I got what your saying. for Cos(x^2) I got -2xsinx^2

Write that as -2xsin(x^2) and I'll be happier. And cos(x^2)^2 still isn't cos(x^4). Two other different things.

Does the same thing apply for Sin (5x) that this is different than Sin 5x?

Is the denominator 2Cos(x^2)?

When in doubt use parentheses. sin 5x could mean either sin(5)*x or sin(5x). Better to be safe than sorry.

meeklobraca said:
Is the denominator 2Cos(x^2)?

Geez. The quotient rule says (u/v)'=(u'*v-v'*u)/v^2. v is cos(x^2). v^2 isn't 2*cos(x^2).

So I got

(5)Cos (5x)Cos(x^2) - Sin (5x)-2xSin(x^2) / 2Cos (x^2)

(x^2) squared is x^4, what is Cos (x^2)^2 then?

(cos(x^2))^2 is different from cos((x^2)^2). One is the square of cos(x^2). The other is cos(x^4). They are different. If you don't believe me pick a number and put it into your calculator. Which do you want?

After doing some review, I got this as a final answer as I don't think I can simplify it anymore.

5Cos(5x)Cos(x^2) + 2x Sin(5x)Sin(x^2) / Cos^2(x^2)

As Dick asked at least a couple of times, use parentheses. Your last answer would be correctly interpreted as
$$5cos(5x)cos(x^2) + \frac{2x sin(5x)Sin(x^2)}{cos^2(x^2)}$$
but I don't think that's what you had in mind.

I see what your saying. I meant for all of that to be divided by Cos^2(x^2)

meeklobraca said:
I see what your saying. I meant for all of that to be divided by Cos^2(x^2)

Then I think you've finally got it.

## 1. What is the derivative of a function?

The derivative of a function is a measure of how the output value of the function changes with respect to the input value. It represents the slope of the function at a specific point.

## 2. Why is finding the derivative important?

Finding the derivative is important because it allows us to analyze the behavior of a function and make predictions about its values. It also helps us solve problems in fields such as physics, engineering, and economics.

## 3. How do you find the derivative of a function?

The derivative of a function can be found using the rules of differentiation, such as the power rule, product rule, and chain rule. These rules involve taking the limit of the difference quotient as the change in the input value approaches zero.

## 4. Can the derivative of a function be negative?

Yes, the derivative of a function can be negative. This means that the function is decreasing at that point, and the slope of the tangent line is negative. A negative derivative can also indicate concave down behavior.

## 5. Is there a difference between derivative and differentiation?

Yes, there is a difference between derivative and differentiation. Derivative refers to the output value of the process of differentiation, which is the instantaneous rate of change of the function at a specific point. Differentiation, on the other hand, is the process of finding the derivative of a function.

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