# Computing derivative (basic calculus question)

1. Feb 28, 2015

### goonking

1. The problem statement, all variables and given/known data
Compute Derivative
y = xx + sin(x)
2. Relevant equations

3. The attempt at a solution
since I have x in the exponent (x^x), I multiply both sides by ln:

ln y = ln xx + ln sin(x)

the x in the exponent comes out into the front, right?

y'/y = x ln x + ln sin (x)

using product rule for xlnx:

y'/y= ((1⋅ln x) + (x) (1/x)) + 1/sin(x) ⋅ cos(x)

y'/y = [(ln x +1) + cos(x)/sin(x)]

multiplying both sides by y to get y' only

y' = [(ln x +1) + cos(x)/sin(x)] ⋅ xx + sin(x)

is this correct? or should the ln xx = Xx ⋅ ln x ???

2. Feb 28, 2015

### SteamKing

Staff Emeritus
In order to take a logarithm of a number, you do not multiply that number by 'ln', just like if you take the sine of an angle, you do not multiply the angle by 'sin'.

The logarithm is a function, just like the sine is a function. Before you dive into calculus, you should learn what functions are.

3. Feb 28, 2015

### SammyS

Staff Emeritus
ln(a + b) ≠ ln(a) + ln(b) .

By the way, that is not multiplying ln times (a + b) .

Hint: Find the derivative of xx separately, by using the natural log (ln) and implicit differentiation. Then include the sine term.

4. Feb 28, 2015

### LCKurtz

In addition to SteamKing's observations, since you already know how to differentiate $\sin x$, just do $x^x$ separately by your method and add it to the derivative of $\sin x$. It will help you avoid the mistakes you made.

5. Feb 28, 2015

### goonking

i saw 2 videos on youtube, apparently you can get the derivative of xx using e and another way without using e, is this correct?

both come out to be xx (lnx+1)

6. Feb 28, 2015

### SammyS

Staff Emeritus
You can express xx in an alternative form using e. This does make it relatively easy to find the derivative .

$\displaystyle x^x = e^{\ln(x^x)}=e^{\,x\,\ln(x)}$

7. Feb 28, 2015

### goonking

ok, I got the new answer now. is it:

y' = (Xx (ln x + 1) + cos(x)) / xx + sin(x)

8. Feb 28, 2015

### SammyS

Staff Emeritus
The trig functions should be completely separate from the xx .

9. Feb 28, 2015

### goonking

what do you mean by that?

10. Feb 28, 2015

### SammyS

Staff Emeritus
You have cos(x) divided by xx .

11. Feb 28, 2015

### goonking

any hint on how to get rid of the trig functions? I have no idea what to do at this point.

12. Mar 1, 2015

### SammyS

Staff Emeritus
You don't get rid of them. They are just not in the same term as xx .

$\left(f(x)+g(x)\right)'=f'(x)+g'(x)$

13. Mar 1, 2015

### goonking

wow, I forgot to multiply both sides by y (the original equation)

then after I do all that, the base and y cancel out and I'm left with X^x (lnx+1) + cos (x))

correct?

14. Mar 1, 2015

### HallsofIvy

i
In addition to what others have said, this is wrong. You have differentiated on the left side but not yet on the right- they are NOT equal!

15. Mar 1, 2015

### SammyS

Staff Emeritus
That is the correct derivative.

Once you find the derivative of xx , you shouldn't need to multiply anything by y .