# Help with new calculus concept

1. Sep 28, 2006

### The Bob

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) $$f(x) = x^{sinh2x}$$

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 realised 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 learnt or cannot find on the internet (and trust me, I have looked).

Any help would be greatly appreciated.

Cheers for now,

2. Sep 28, 2006

### matt grime

It's the chain rule. You've met the chain rule?

3. Sep 28, 2006

### neutrino

$$\log_{e}f = ?$$

4. Sep 28, 2006

### The Bob

Is it truely 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,

5. Sep 28, 2006

### neutrino

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. Sep 28, 2006

### The Bob

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

$$f'(x) = 2 cosh2x . sinh2x . x^{(sinh2x - 1)}$$

?

Cheers,

7. Sep 28, 2006

### The Bob

But then I have a $$log_e f'(x)$$, do I not?

8. Sep 28, 2006

### The Bob

Ok, so will I have

$$\frac{d(log_e y)}{dx} = \frac{sinh2x}{x} + (2cosh2x)(log_e x)$$

?

9. Sep 28, 2006

### neutrino

$$\frac{d(\log f(x))}{dx} = \frac{1}{f(x)}f'(x)$$

This is yet another application of the chain rule...

10. Sep 28, 2006

### neutrino

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

11. Sep 28, 2006

### The Bob

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

12. Sep 28, 2006

### matt grime

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. Sep 28, 2006

### neutrino

But in the OP's case, the base is not e.

14. Sep 28, 2006

### matt grime

Doh. stupid me.

15. Sep 28, 2006

### Data

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

$$f(x) = e^{\sinh (2x)\ln{x}}$$

and apply the chain rule directly.

16. Sep 28, 2006

### The Bob

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

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

$$f(x) = e^{\sinh (2x)\ln{x}}$$ Let u = sinh(2x).ln(x)

$$\Rightarrow f(x) = y = e^u$$

$$\Rightarrow f'(x) = \frac{dy}{dx} = e^u$$

Chain rule says that $$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$

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

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

$$\Rightarrow \frac{dy}{dx} = x^{\sinh (2x)} \times (ln(x) \cdot 2cosh(2x) \ + \ \frac{sinh(2x)}{x})$$

Is this now correct?

Cheers,

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: Sep 28, 2006
17. Sep 28, 2006

### neutrino

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. Sep 29, 2006

### The Bob

Aha.... ok. 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 text book I have, will yield more answers before I embarasse myself here.

Thanks again everyone,