How to differentiate kx^(kx)?

[Captain Sunshine]

Okay, so I've been mulling this one over for a couple of days now, but haven't come to a satisfactory answer. How does one find the derivative of an equation such as y = 3x^3x logarithmically?

The answer given is: 9x^3x (ln x + 1)

My best attempt was to take the natural log of both sides and solve for the derivative:


ln y = 3 • ^d/_dx 3x ln x

¹/_y • ^dy/_dx = 3 (3 ln x + 3)

^dy/_dx = 9 • y (ln x + 1) = 27x^3x(ln x + 1)

No good.

I looked around and discovered I could get the correct answer like this:

y = 3 • ^d/_dx e^{(3x ln x)}

(Let u = 3x ln x)

y = 3 • ^d/_due^{u d}/_dx 3x ln x

y = 3 e^u (3 ln x + 3)

y = 9x^3x (ln x + 1)


Okay, so I have a way of solving this. However, on the worksheet (Question 3: http://cdn.kutasoftware.com/Worksheets/Calc/03 - Logarithmic Differentiation.pdf) the method shown for solving the equation most resembles my failed attempt.

Could someone please point me in the correct direction with this? I'm sure I must be misusing a log property somewhere.

Thanks.

 

[stapel]

How does one find the derivative of an equation such as y = 3x^3x logarithmically?

The answer given is: 9x^3x (ln x + 1)

My best attempt was to take the natural log of both sides and solve for the derivative:

I think this would be the standard method...

ln y = 3 • ^d/_dx 3x ln x

...but this is not how it's done.

Instead, try working one step at a time. (Logging one side while, I think, differentiating the other is not going to lead anywhere happy.)

Start with the original equation:

. . . . .\(\displaystyle y\, =\, 3x^{3x}\)

Log each side:

. . . . .\(\displaystyle \ln(y)\, =\, \ln\left(3x^{3x}\right)\)

Apply log rules:

. . . . .\(\displaystyle \ln(y)\, =\, \ln(3)\, +\, \ln\left(x^{3x}\right)\)

. . . . .\(\displaystyle \ln(y)\, =\, \ln(3)\, +\, (3x)\, \ln(x)\)

Differentiate, using the Product Rule on the second term on right-hand side:

. . . . .\(\displaystyle \dfrac{1}{y}\, \dfrac{dy}{dx}\, =\, (0)\, \dfrac{dx}{dx}\, +\, \bigg(\, (3)\,\dfrac{dx}{dx}\, \ln(x)\, +\, (3x)\, \dfrac{1}{x}\,\dfrac{dx}{dx}\,\bigg)\)

Simplify:

. . . . .\(\displaystyle \dfrac{1}{y}\, \dfrac{dy}{dx}\, =\, 3\, \ln(x)\, \dfrac{dx}{dx}\, +\, 3\, \dfrac{dx}{dx}\)

Note that y = 3x^3x and dx/dx = 1, so:

. . . . .\(\displaystyle \dfrac{1}{3x^{3x}}\, \dfrac{dy}{dx}\, =\, 3\, \ln(x)\, +\, 3\)

Multiply through:

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, (3)\, \left(3x^{3x}\right)\, \ln(x)\, +\, (3)\, \left(3x^{3x}\right)\)

You can probably take things from there.

 

[Ishuda]

Okay, so I've been mulling this one over for a couple of days now, but haven't come to a satisfactory answer. How does one find the derivative of an equation such as y = 3x^3x logarithmically?

The answer given is: 9x^3x (ln x + 1)

My best attempt was to take the natural log of both sides and solve for the derivative:


ln y = 3 • ^d/_dx 3x ln x <========== WRONG

¹/_y • ^dy/_dx = 3 (3 ln x + 3)

^dy/_dx = 9 • y (ln x + 1) = 27x^3x(ln x + 1)

No good.

I looked around and discovered I could get the correct answer like this:

y = 3 • ^d/_dx e^{(3x ln x)}

(Let u = 3x ln x)

y = 3 • ^d/_due^{u d}/_dx 3x ln x

y = 3 e^u (3 ln x + 3)

y = 9x^3x (ln x + 1)


Okay, so I have a way of solving this. However, on the worksheet (Question 3: http://cdn.kutasoftware.com/Worksheets/Calc/03 - Logarithmic Differentiation.pdf) the method shown for solving the equation most resembles my failed attempt.

Could someone please point me in the correct direction with this? I'm sure I must be misusing a log property somewhere.

Thanks.

See red above. Should be
ln y = ln (3 x^(3x)) = ln(3) + ln(x^(3x))
and continue from there.

 

[ksdhart2]

The error comes in your first step, when you take the logarithm of both sides. You're attempting to isolate the scalar multiple of 3 from the x^3x term, then take the log of only the x^3x term, which is not allowed.If you take the logarithm of one side of the equation, you must take the logarithm of the entirety of the other side, in order for the equation to hold true. To use some real numbers, let's say x = 1. Your incorrect equation says that:

\(\displaystyle y=3x^{3x}=3^3=27\)

\(\displaystyle \ln{y}=3 \cdot \ln{x^{3x}} \implies ln{27} = 3 \cdot \ln{1^3} \implies \ln{27}=3 \cdot \ln{1} \implies 3.29=0\)

Oops. On the other hand, the correct equation says:

\(\displaystyle \ln{y}=\ln{3x^{3x}} \implies ln{27} = \ln{3^3} \implies \ln{27}=\ln{27}\)

Now that checks out.

 

[Captain Sunshine]

Ah, thank you everyone. And for such fast replies too.

Such a silly thing to overlook! Once I saw it pointed out, I realised what a glaring error it was. I have to say, I'm a little bit ashamed of myself for not seeing it straight away

Well, thanks again guys. I really appreciate the help, and now I can relax