# Solving Sec Homework: Differentiate Ln(cos(5x)) wrt x

• Roodles01
In summary, the conversation discusses a student's attempt at differentiating Ln(cos(5x)), their use of the chain rule and standard derivatives, and the different method and answer provided by WolframAlpha. The student also asks for clarification on the relationship between sec(w) and tan(w).
Roodles01

## Homework Statement

Having started to differentiate Ln(cos(5x)) wrt x I checked the answer with WolframAlpha & got a different method & answer too.

differentiate
ln(cos(5x))

## The Attempt at a Solution

I used the Chain rule d(ln(cos(5x)))/dx = d(ln(u))/du * du/dx
Where u = cos(5x)

d(ln(u))/du = 1/u which I have in a table of standard derivatives
= 1/(cos(5x)

du/dx = -sin(5x) again, from a table of standard derivatives.

So according to me d(ln(cos(5x)))/dx = 1/cos(5x) * -sin(5x)

WolframAlpha goes a different route & has the answer
-5 tan(5x)

I see how this differs from mine so;
1. why am I wrong,
2. how did WA get from sec(5x)(sin(5x)(-(d(5x)/dx))) to -5tan(5x). i.e how is tan defined in terms of sec?
Sorry to be dense!

Last edited:
Roodles01 said:
u = cos(5x)
du/dx = -sin(5x)

Something wrong here~

and sec x = 1 / cos x

Hmmm!
My table of standard derivatives shows;
if function f(x) = cos(ax),
the derivative f'(x) = -a sin(ax)

In my example a = 5, so -5 sin(5x)

Ah! I see the bit I missed out.

Last edited:
Roodles01 said:

## Homework Statement

Having started to differentiate Ln(cos(5x)) wrt x I checked the answer with WolframAlpha & got a different method & answer too.

differentiate
ln(cos(5x))

## The Attempt at a Solution

I used the Chain rule d(ln(cos(5x)))/dx = d(ln(u))/du * du/dx
Where u = cos(5x)

d(ln(u))/du = 1/u which I have in a table of standard derivatives
= 1/(cos(5x)

du/dx = -sin(5x) again, from a table of standard derivatives.

So according to me d(ln(cos(5x)))/dx = 1/cos(5x) * -sin(5x)

WolframAlpha goes a different route & has the answer
-5 tan(5x)

I see how this differs from mine so;
1. why am I wrong,
2. how did WA get from sec(5x)(sin(5x)(-(d(5x)/dx))) to -5tan(5x). i.e how is tan defined in terms of sec?
Sorry to be dense!

What is the *definition* of tan(w)? Surely your textbook tells you that! If not, try Google.

RGV

tan(w) = sin(w)/cos(w)
sec(w) = 1/cos(w)

I haven't yet found a relationship between sec (w) & tan(w) which also involves multiplying it by sin(w).
From the WA page I can, of course, see that it has the relationship as tan(w) = sec(w)sin(w). I would like someone to confirm this, please & say whether this is a standard which I haven't yet encountered.

Last edited:
You have it right there- you originally gave, as the derivative, (1/cos(5x))(-sin(5x)). You now recognize that, because the derivative of 5x is 5, it should be (5)(1/cos(5x)(-sin(5x))= -5(sin(5x)/cos(5x))= -5tan(5x).

Amazingly easy when pointed out.
. . . 'wood for the trees - again!

Completed my original question
f(x)=sec(5x)ln(cos(5x))
after product rule & then chain rule for different bits.
Thank you all.

## 1. What is the basic concept of differentiation?

Differentiation is a mathematical process used to find the rate of change of a function with respect to its independent variable. It involves finding the derivative of a function, which represents the slope of the tangent line at a specific point on the function.

## 2. How do I differentiate a natural logarithmic function?

To differentiate a natural logarithmic function, you can use the chain rule. First, identify the inner function, in this case cos(5x). Then, take the derivative of the inner function and multiply it by the derivative of the outer function, which is ln(x). In this case, the derivative of cos(5x) is -5sin(5x), so the final answer is -5sin(5x)/x.

## 3. What are the steps for differentiating a composite function?

The steps for differentiating a composite function are as follows:

1. Identify the inner function and the outer function.
2. Use the chain rule to take the derivative of the inner function.
3. Multiply the derivative of the inner function by the derivative of the outer function.

## 4. How do I differentiate a trigonometric function?

To differentiate a trigonometric function, you can use the chain rule or the trigonometric identities. For example, to differentiate cos(x), you can use the identity sin(x) = cos(x + π/2) to rewrite the function as sin(x + π/2) and then use the chain rule to find the derivative.

## 5. What is the purpose of differentiating a function?

The main purpose of differentiating a function is to find the slope of the tangent line at a specific point on the function. This can be used to determine the rate of change of the function, find critical points, and analyze the behavior of the function. It is also a fundamental concept in calculus and is used in many real-world applications such as physics, economics, and engineering.

Replies
2
Views
1K
Replies
15
Views
1K
Replies
5
Views
2K
Replies
2
Views
2K
Replies
4
Views
4K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
7
Views
2K
Replies
3
Views
1K
Replies
7
Views
1K