Computing derivative (basic calculus question)

In summary: You should be able to find the derivative of sin(x) without multiplying anything by y . And then you add the two together to get the complete derivative.You have some arithmetic errors in your work above, but I think you understand the calculus involved.The derivative of xx is xx(lnx + 1). The derivative of sin(x) is cos(x). So the complete derivative is xx(lnx+1)+cos(x)In summary, the derivative of y = xx + sin(x) is xx(lnx+1) + cos(x). The derivative of xx is xx(lnx + 1) and the derivative of sin(x) is cos(x). The complete derivative is found by adding the derivatives of each term, which gives xx
  • #1
goonking
434
3

Homework Statement


Compute Derivative
y = xx + 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 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 ?
 
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  • #2
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
goonking said:

Homework Statement


Compute Derivative
y = xx + 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 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 ?
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
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.
 
  • #5
SammyS said:
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.
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
goonking said:
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)
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
ok, I got the new answer now. is it:

y' = (Xx (ln x + 1) + cos(x)) / xx + sin(x)
 
  • #8
goonking said:
ok, I got the new answer now. is it:

y' = (Xx (ln x + 1) + cos(x)) / xx + sin(x)
The trig functions should be completely separate from the xx .
 
  • #9
SammyS said:
The trig functions should be completely separate from the xx .
what do you mean by that?
 
  • #10
goonking said:
what do you mean by that?
You have cos(x) divided by xx .
 
  • #11
SammyS said:
You have cos(x) divided by xx .
any hint on how to get rid of the trig functions? I have no idea what to do at this point.
 
  • #12
goonking said:
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 xx .

##\left(f(x)+g(x)\right)'=f'(x)+g'(x) ##
 
  • #13
SammyS said:
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) ##
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
i
goonking said:

Homework Statement


Compute Derivative
y = xx + 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 xx + 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)] ⋅ xx + sin(x)
is this correct? or should the ln xx = Xx ⋅ ln x ?
 
  • #15
goonking said:
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 xx , you shouldn't need to multiply anything by y .
 

FAQ: Computing derivative (basic calculus question)

What is a derivative in calculus?

A derivative in calculus is a measure of how a function changes as its input changes. It represents the instantaneous rate of change of a function at a specific point.

How do you compute a derivative?

To compute a derivative, you need to use the rules of differentiation, which involve taking the limit of the difference quotient as the change in the input approaches zero. This is typically done using algebraic manipulation and the application of various derivative rules.

What is the difference between a derivative and a derivative function?

A derivative represents the slope of a tangent line to a function at a specific point, while a derivative function represents the set of all derivatives of a function at every point in its domain. Essentially, the derivative function is a collection of all the slopes of tangent lines for a given function.

Why is the derivative important?

The derivative is important because it allows us to analyze the rate of change of a function, which has many practical applications in fields such as physics, engineering, economics, and more. It also helps us find the maximum and minimum values of a function, which is useful for optimization problems.

Can you give an example of computing a derivative?

Sure, let's say we have the function f(x) = 3x^2 + 5x. To find the derivative of this function, we use the power rule, which states that the derivative of x^n is nx^(n-1). So, using this rule, we can find that the derivative of f(x) is f'(x) = 6x + 5. This represents the instantaneous rate of change of f(x) at any given point.

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