Computing derivative (basic calculus question)

[goonking]

Homework Statement

Compute Derivative
y = x^x + sin(x)

Homework Equations

The Attempt at a Solution

since I have x in the exponent (x^x), I multiply both sides by ln:

ln y = ln x^x + 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)] ⋅ x^x + sin(x)
is this correct? or should the ln x^x = X^x ⋅ ln x ?

 

[SteamKing]

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.

 

[SammyS]

Homework Statement

Compute Derivative
y = x^x + sin(x)

Homework Equations

The Attempt at a Solution

since I have x in the exponent (x^x), I multiply both sides by ln:

ln y = ln x^x + 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)] ⋅ x^x + sin(x)
is this correct? or should the ln x^x = X^x ⋅ ln x ?

ln(a + b) ≠ ln(a) + ln(b) .

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

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

 

[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.

[Edit] Didn't see Sammy's reply.

 

[goonking]

ln(a + b) ≠ ln(a) + ln(b) .

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

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

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

both come out to be x^x (lnx+1)

 

[SammyS]

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

both come out to be x^x (lnx+1)

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

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

 

[goonking]

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

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

 

[SammyS]

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

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

The trig functions should be completely separate from the x^x .

 

[goonking]

The trig functions should be completely separate from the x^x .

what do you mean by that?

 

[SammyS]

what do you mean by that?

You have cos(x) divided by x^x .

 

[goonking]

You have cos(x) divided by x^x .

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

 

[SammyS]

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

You don't get rid of them. They are just not in the same term as x^x .

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

 

[goonking]

You don't get rid of them. They are just not in the same term as x^x .

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

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?

 

[HallsofIvy]

i

Homework Statement

Compute Derivative
y = x^x + sin(x)

Homework Equations

The Attempt at a Solution

since I have x in the exponent (x^x), I multiply both sides by ln:

ln y = ln x^x + ln sin(x)

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

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

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!

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)] ⋅ x^x + sin(x)
is this correct? or should the ln x^x = X^x ⋅ ln x ?

 

[SammyS]

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?

That is the correct derivative.

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