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,