# Thread: Making some progress on differentiation problem

1. ## 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!

2. 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?

$y = x^{3x}$

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

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

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

You aren't far off it. Just use the product rule on the RHS.

3. 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.

4. 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!

5. 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. $y = x^{3x}$

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

So multiply through by $x^{3x}$ to get your equation $dy/dx =$ purely in terms of x.

6. I submit the following and it shows up as incorrect..

I may be missing something?

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

7. 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 $2x + 3 \ln(x)$ is NOT the derivative of $3x. \ln {x}$

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

$y = x^{3x}$

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

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

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

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

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

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

$y = x^{3x}$

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

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

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

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

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

still not correct

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

still not correct
That's right.

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

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

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