# Making some progress on differentiation problem

• Oct 26th 2009, 07:36 PM
frozenflames
Making some progress on differentiation problem
f(x) = x^(3x)

So far i used implicit differentiation and get
(1/y)(y')=2x + 3ln(x)

but I am way off,

any help would be appreciated!
• Oct 26th 2009, 07:38 PM
Mush
Quote:

Originally Posted by frozenflames
f(x) = x^(3x)

So far i used implicit differentiation and get
(1/y)(y')=2x + 3ln(x)

but I am way off,

any help would be appreciated!

Why are you way off?

$\displaystyle y = x^{3x}$

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

$\displaystyle \ln(y) = 3x.\ln(x)$

$\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} 3x.\ln(x)$

You aren't far off it. Just use the product rule on the RHS.
• Oct 26th 2009, 07:39 PM
rn443
Quote:

Originally Posted by frozenflames
f(x) = x^(3x)

So far i used implicit differentiation and get
(1/y)(y')=2x + 3ln(x)

but I am way off,

any help would be appreciated!

Express y as e^(3x*ln(x)) and use the chain rule.
• Oct 26th 2009, 07:42 PM
frozenflames
Quote:

Originally Posted by rn443
Express y as e^(3x*ln(x)) and use the chain rule.

i'm not sure how to solve for y'....because I would assume I would multiply both sides by "y" to get the result but the solution indicates that "y" is not supposed to be in the solution at all! :(
• Oct 26th 2009, 07:44 PM
Mush
Quote:

Originally Posted by frozenflames
i'm not sure how to solve for y'....because I would assume I would multiply both sides by "y" to get the result but the solution indicates that "y" is not supposed to be in the solution at all! :(

Yes, but you know what y is from the original equation. Simply substitute it in. $\displaystyle y = x^{3x}$

So $\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^{3x}} \frac{dy}{dx}$

So multiply through by $\displaystyle x^{3x}$ to get your equation $\displaystyle dy/dx =$ purely in terms of x.
• Oct 26th 2009, 08:08 PM
frozenflames
I submit the following and it shows up as incorrect..

I may be missing something?

x^(3x)[2x + 3ln(x)]
• Oct 26th 2009, 08:13 PM
Mush
Quote:

Originally Posted by frozenflames
I submit the following and it shows up as incorrect..

I may be missing something?

x^(3x)[2x + 3ln(x)]

Yes. As I said before $\displaystyle 2x + 3 \ln(x)$ is NOT the derivative of $\displaystyle 3x. \ln {x}$

You have carried out your differentiation incorrectly on the RHS of the equation AFTER you've taken the natural logarithm of both sides.

$\displaystyle y = x^{3x}$

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

$\displaystyle \ln(y) = 3x \ln (x)$

$\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3x . \ln(x))$

You must use the product rule on the RHS to differentiate.
• Oct 26th 2009, 08:42 PM
frozenflames
Quote:

Originally Posted by Mush
Yes. As I said before $\displaystyle 2x + 3 \ln(x)$ is NOT the derivative of $\displaystyle 3x. \ln {x}$

You have carried out your differentiation incorrectly on the RHS of the equation AFTER you've taken the natural logarithm of both sides.

$\displaystyle y = x^{3x}$

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

$\displaystyle \ln(y) = 3x \ln (x)$

$\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3x . \ln(x))$

You must use the product rule on the RHS to differentiate.

that gives me this: 3x(1/x)+3ln(x)

still not correct
• Oct 26th 2009, 08:45 PM
Mush
Quote:

Originally Posted by frozenflames
that gives me this: 3x(1/x)+3ln(x)

still not correct

That's right.

$\displaystyle \frac{1}{y} \frac{dy}{dx} = 3 + 3\ln(x)$

$\displaystyle \frac{dy}{dx} = (3+3\ln(x)) \times y$

$\displaystyle \frac{dy}{dx} = 3x^{3x} (1+\ln(x))$