Help with new calculus concept

In summary, the conversation revolves around a student struggling with a question involving finding the derivative of a function with a base of sinh(2x) and an exponent of ln(x). The group provides guidance and suggests using the chain rule to solve the problem. The student eventually arrives at the correct solution with the help of the group.
  • #1
The Bob
1,126
0
Hi all (it has been a while again),

I have just started Universtiy and after my second lecture I was given a question sheet to complete in two weeks. I have completed 14 out of the 16 questions set, 2 of which I just haven't done yet and another will require a little more thought before I need some help.

The question I am having a problem with is this one:

4. Find f'(x) for each of the following f.
(d) [tex] f(x) = x^{sinh2x} [/tex]

I would give some sort of attempt, like you all expect, before I get any help but I am afraid I have never come across this sort of equation before. I really do not know where to start. I did think about letting u = sinh2x and doing a sustitution but I realized that would not work so I am really stuck. I can but assume I need some sort of rule or theorem that I have either not learned or cannot find on the internet (and trust me, I have looked).

Any help would be greatly appreciated.

Cheers for now,

The Bob (2004 ©)
 
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  • #2
It's the chain rule. You've met the chain rule?
 
  • #3
[tex]\log_{e}f = ?[/tex]
 
  • #4
Is it truly as simple as the chain rule?

Also I have never done a f'(x) question by using log base e before I start.

So even if it is as simple as the chain rule (which has caused myself to be quiet embarrassed) what does the right hand side become? I am not going to pretend I understand at this stage, despite feeling I should be able to.

Cheers,

The Bob (2004 ©)
 
  • #5
Use the rule log(a^b) = b log(a), then differentiate the equation, using the the product and chain rules, with respect to x.
 
  • #6
Let me have a quick go.

Does sinhx need to be substituted for some lettter (e.g. u)? Then I can differentiate with respect to u (for f(x)) and with respect to x (for u).

So is

[tex]f'(x) = 2 cosh2x . sinh2x . x^{(sinh2x - 1)}[/tex]

?

Cheers,

The Bob (2004 ©)
 
  • #7
neutrino said:
Use the rule log(a^b) = b log(a), then differentiate the equation, using the the product and chain rules, with respect to x.

But then I have a [tex]log_e f'(x)[/tex], do I not?

The Bob (2004 ©)
 
  • #8
Ok, so will I have

[tex] \frac{d(log_e y)}{dx} = \frac{sinh2x}{x} + (2cosh2x)(log_e x)[/tex]

?

The Bob (2004 ©)
 
  • #9
[tex]\frac{d(\log f(x))}{dx} = \frac{1}{f(x)}f'(x)[/tex]

This is yet another application of the chain rule...
 
  • #10
The Bob said:
[tex]f'(x) = 2 cosh2x . sinh2x . x^{(sinh2x - 1)}[/tex]

You can't use the same rule for differentiating x^n, with n constant. sinh(2x) is a function of x.
 
  • #11
neutrino said:
You can't use the same rule for differentiating x^n, with n constant. sinh(2x) is a function of x.

I did think as much but felt it was, stupidly, worth a try. :frown:

The Bob (2004 ©)
 
  • #12
Forget logs. There is no need to use them.

The derivative of exp(x) is exp(x). The derivative of sinh(2x) is 2cosh(2x). Just apply the chain rule

(f(g(x)))' = g'(x)f'(g(x)).
 
  • #13
matt grime said:
Forget logs. There is no need to use them.

The derivative of exp(x) is exp(x). The derivative of sinh(2x) is 2cosh(2x). Just apply the chain rule

(f(g(x)))' = g'(x)f'(g(x)).
But in the OP's case, the base is not e.
 
  • #14
Doh. stupid me.
 
  • #15
You can either take the log of both sides (as suggested by others), or just rewrite it as

[tex]f(x) = e^{\sinh (2x)\ln{x}}[/tex]

and apply the chain rule directly.
 
  • #16
matt grime said:
Doh. stupid me.

I was wondering why you said to just use the chain rule.

Data said:
You can either take the log of both sides (as suggested by others), or just rewrite it as

[tex]f(x) = e^{\sinh (2x)\ln{x}}[/tex]

and apply the chain rule directly.

I can see how, from this, I can get a solution:

[tex]f(x) = e^{\sinh (2x)\ln{x}}[/tex] Let u = sinh(2x).ln(x)

[tex]\Rightarrow f(x) = y = e^u[/tex]

[tex]\Rightarrow f'(x) = \frac{dy}{dx} = e^u[/tex]


Chain rule says that [tex]\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}[/tex]

[tex]\frac{du}{dx}[/tex] requires the product rule so [tex]\frac{du}{dx} = ln(x) \cdot 2 \cosh (2x) \ + \ \frac{\sinh (2x)}{x}[/tex]

Thus [tex]\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = (e^{\sinh (2x)\ln{x}}) \times (ln(x) \cdot 2 \cosh (2x) \ + \ \frac{\sinh (2x)}{x})[/tex]

[tex]\Rightarrow \frac{dy}{dx} = x^{\sinh (2x)} \times (ln(x) \cdot 2cosh(2x) \ + \ \frac{sinh(2x)}{x})[/tex]

Is this now correct?

Cheers,

The Bob (2004 ©)

P.S. It would appear that the LaTex is not working too well so here is the final equation in text:

dy/dx = (x^(sinh(2x))) x ((ln(x)).(2cosh(2x)) + (sinh(2x)/x))
 
Last edited:
  • #17
Apart from the line where you said, f'(x) = dy/dx = e^u, you've are correct. It should've been e^u.du/dx, but since you got chain rule part right, I'll let you off. ;)
 
  • #18
neutrino said:
Apart from the line where you said, f'(x) = dy/dx = e^u, you've are correct. It should've been e^u.du/dx, but since you got chain rule part right, I'll let you off. ;)

Aha... ok. :biggrin: Thanks for everyone's help. I must admit that was a technique (rather than a principle or theorem) that I had not done before, as simple as it was.

I do hope in future the interent, or the large textbook I have, will yield more answers before I embarasse myself here.

Thanks again everyone,

The Bob (2004 ©)
 

1. What is calculus?

Calculus is a branch of mathematics that deals with the study of change and its relationship to different quantities. It is divided into two main branches: differential calculus, which focuses on the rate of change of a function, and integral calculus, which deals with the accumulation of quantities over a given interval.

2. What is a new calculus concept?

A new calculus concept refers to a mathematical idea or principle that is not commonly known or understood in the field of calculus. It could be a new technique, formula, or approach to solving problems within the study of calculus.

3. Why is it important to learn new calculus concepts?

Learning new calculus concepts is important because it allows us to expand our understanding of the subject and apply it to real-world problems. It also helps us to improve our problem-solving skills and think critically about mathematical concepts.

4. How can I better understand a new calculus concept?

To better understand a new calculus concept, it is important to start by reviewing the fundamentals and building a strong foundation. You can also try practicing problems and seeking help from a teacher or tutor. Visual aids and online resources can also be helpful in understanding new concepts.

5. What are some common challenges when learning new calculus concepts?

Some common challenges when learning new calculus concepts include understanding the language and notation used, grasping abstract concepts, and applying the concepts to real-world problems. It is important to be patient and persistent, and to seek help when needed.

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