Finding the Derivative of Exponential Functions

[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)

 

[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)?

 

[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:
$$\frac{dh}{dx}=(\frac{\partial{H}}{\partial{u}}\frac{dU}{dx}+\frac{\partial{H}}{\partial{v}}\frac{dV}{dx}+\frac{\partial{H}}{\partial{w}}\frac{dW}{dx})\mid_{(u,v,w)=(U(x),V(x),W(x)}$$

 

[thomasrules]

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

x^x(1+lnx)

 

[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

 

[dextercioby]

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

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.

 

[HallsofIvy]

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

x^x(1+lnx)

Yes, that's perfectly correct.

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

Also correct. In particular, if y= x^x then ln y= x ln x so
(1/y)y'= ln x+ x(1/x)= ln(x)+ 1. y'= y(ln x+ 1)= x^x(ln x+ 1) as thomasrules said. You can use (x^x)'= x^x(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.

 

[arildno]

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

x^x(1+lnx)

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)):
$$\frac{dy}{dx}=x*x^{x-1}+x^{x}\ln(x)=x^{x}(1+\ln(x))$$

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

As noted, there are various valid methods to do this correctly. Logarithmic differentiation is ONE technique, among several.