# Exponential Derivative

1. Nov 22, 2006

### thomasrules

How do you find the derivative of:

y=x(3x)^x^2
?

I though it was yprime=(1)3x^x^2+x(3x^x^2)ln(1+3x)(2x)

2. Nov 22, 2006

### JoAuSc

I got

1*3x^x^2+x(3x^x^2)*(2x ln(3x) + x).

I'm guessing you just did some bad algebra. Maybe instead of ln(1+3x) you meant (ln(3x) + 1)?

3. Nov 22, 2006

### arildno

You could do it in various ways.
One method (ultimately justified by the multi-variable chain rule) is to regard one x as varying, the others constant, and then add the results:
$$\frac{dy}{dx}=(3x)^{x^{2}}+3x^{3}(3x)^{x^{2}-1}+2x^{2}(3x)^{x^{2}}\ln(3x)$$

To do this properly, we may define:
$$H(u,v,w)=uv^{w}, U(x)=x, V(x)=3x, W(x)=x^{2}$$
Then, we may define the function:
$$h(x)=H(U(x),V(x),W(x))$$
and we have, by the multi-variable chain rule:
$$\frac{dh}{dx}=(\frac{\partial{H}}{\partial{u}}\frac{dU}{dx}+\frac{\partial{H}}{\partial{v}}\frac{dV}{dx}+\frac{\partial{H}}{\partial{w}}\frac{dW}{dx})\mid_{(u,v,w)=(U(x),V(x),W(x)}$$

Last edited: Nov 22, 2006
4. Nov 22, 2006

### thomasrules

i thought the law of y=x^x was:

x^x(1+lnx)

5. Nov 22, 2006

### stunner5000pt

i was taught taht you should find the log of both sides and differentiate from there

so then you question
$$y = x(3x)^{x^2}$$
will turn into

$$\ln y = \ln (x(3x)^{x^2})$$

and differentiate both sides from here

6. Nov 23, 2006

### dextercioby

There's a name for what you did: logarithmic differentiation. It basically allows one to get the derivative of any (no matter how complicated) elementary function.

Daniel.

7. Nov 23, 2006

### HallsofIvy

Staff Emeritus
Yes, that's perfectly correct.

Also correct. In particular, if y= xx then ln y= x ln x so
(1/y)y'= ln x+ x(1/x)= ln(x)+ 1. y'= y(ln x+ 1)= xx(ln x+ 1) as thomasrules said. You can use (xx)'= xx(ln x+ 1) together with the product and chain rules to do this more complicated version:
$$y= x(3x)^{x^2}$$
then
$$y'= (3x)^{x^2}+ x((3x)^{x^2})(ln(3x)+ 1)(2x)$$
or you can say
$$\ln y = \ln (x(3x)^{x^2})= ln x+ x^2 ln 3+ x^2 ln(x)$$
and differentiate that.

8. Nov 23, 2006

### arildno

Sure enough, and you can do find this out in various ways:
1. One variable chain rule
$$y=x^{x}=e^{x\ln(x)}=e^{u(x)},\frac{dy}{dx}=\frac{d}{du}e^{u}*\frac{du}{dx}=e^{u(x)}*(\ln(x)+x*\frac{1}{x})=x^{x}(1+\ln(x))$$
2. Logarithmic differentiation:
$$Y(x)=\ln(y(x))=x\ln(x),\frac{dY}{dx}=\ln(x)+1=\frac{dY}{dy}\frac{dy}{dx}\to\frac{1}{y(x)}\frac{dy}{dx}=1+\ln(x)\to\frac{dy}{dx}=y(x)(1+\ln(x))=x^{x}(1+\ln(x))$$
3. Multivariable chain rule (regard one x as constant, the others varying, then add the results)):
$$\frac{dy}{dx}=x*x^{x-1}+x^{x}\ln(x)=x^{x}(1+\ln(x))$$
As noted, there are various valid methods to do this correctly. Logarithmic differentiation is ONE technique, among several.