# Derivative of log

1. Feb 13, 2009

### ldbaseball16

1. The problem statement, all variables and given/known data
find f '(x) of 5^xsinhx

2. Relevant equations
i feel like im missing somthing can someone direct me to the right direction pleasee???

3. The attempt at a solution
product rule (f ')(g)+(f)(g ') f=5^x g=sinhx

5^xsinhx= e^ln(5)^xsinhx

(e^ln(5)^x)(sinhx)+(5^x)(coshx)(ln5)

(e^ln(5)^x)(sinhx)+(5^x)(ln5)(coshx)??????

2. Feb 13, 2009

### nicksauce

Well it looks like you took the derivative of none of the terms in the first term, and both of the terms in the second term. Try to fix that.

3. Feb 13, 2009

### Unco

The factor of ln(5) is in the wrong term.

Write them out separately:
f =
f' =
g =
g' =

Then put them together.

By the way, ensure you use parentheses to remove ambiguity when typing exponents; e.g., 5^x = (e^(ln(5)))^x = $$\left(e^{\ln{5}}\right)^x$$ etc.

4. Feb 13, 2009

### ldbaseball16

"none of the terms in the first term" what do you mean by this?

5. Feb 14, 2009

### Staff: Mentor

As Unco pointed out, your expression above is ambiguous. It's not clear what the exponent on 5 is. It could be x, or it could be x*sinh(x). In other words, is this function (5^x)* sinh(x) or is it (5^x)^sinh(x)?
Now from this work it appears that the function is (5^x) * sinh(x), in other words, as the product of two functions. Inasmuch as you're using the product rule, you must think that the function is (5^x)*sinh(x).
Now it looks like an exponential function raised to a power, so the function is not a product, which means that the product rule does not apply.

What exactly is the function? We can't help you if you don't know what it is.

6. Feb 14, 2009

### HallsofIvy

Staff Emeritus
5^x is NOT "e^ln(5)^x" it is e^(x ln(5))

Which means that (5^x)'= (e^(x ln(5)))'= ln(5) e^(x ln(5))= ln(5) 5^x.

7. Feb 14, 2009

### ldbaseball16

here is the original problem find f '(x).

f(x)=5^xSinhx does this clarify???

8. Feb 14, 2009

### tiny-tim

oooh, why can't everyone learn how to use the X2 tag (just above the Reply box)?

It's e(ln5)xsinhx.​

9. Feb 16, 2009

### ldbaseball16

yea and thats what i put on top??? so since the product rule is (f ')(g)+(f)(g ')
so wouldnt the answer be (e^ln(5)xsinhx+(5^x)(coshx)?????

10. Feb 16, 2009

### tiny-tim

nooo! try the X2 tag!
product rule?

what product?

use the chain rule first!

11. Feb 16, 2009

### ldbaseball16

i don't think its the right answer e^sinhx(coshx)

Last edited: Feb 16, 2009
12. Feb 17, 2009

### ldbaseball16

ok so i got 5^xcoshx+5^x(ln5)sinhx?????????????????

13. Feb 17, 2009

### tiny-tim

not even close

show us your full working …

differentiate e(ln5)xsinhx using the chain rule

14. Feb 17, 2009

### ldbaseball16

i used the chain rule and i got e^(ln5xsinhx)(ln5)???????

15. Feb 17, 2009

### tiny-tim

what is the derivative of (ln5)xsinhx?

16. Feb 17, 2009

### ldbaseball16

1/5xcoshx

17. Feb 17, 2009

### Staff: Mentor

It looks like you are doing this: d/dx(ln 5) = 1/5

If so, that's wrong.

Also, when you differentiate (ln 5)* x*sinh(x), you should be getting two terms.

The answer in your previous post might be interpreted in several ways:
$$\frac{1}{5x cosh(x)}$$
$$\frac{1}{5x}cosh(x)$$
$$\frac{1}{5}xcosh(x)$$

Use parentheses or learn LaTex!

18. Feb 17, 2009

### ldbaseball16

ln5*sinh(x)+ln5*x*cosh(x)?????????

19. Feb 17, 2009

### Staff: Mentor

Yes, that's it. Now you can respond to Tiny Tim's request in post 13.