Differentiation With 'x' Both in Base and in Exponent

[modulus]

I have just begun studying differentiaition and I was getting confused with how to differentiate a function of x in which x is in the base and is in the exponent as well (for example, x^2x) with respect to a=hte cahnge in 'x'.
I remember my teacher telling me something about applying log, but, I got kinda drowsy... I wasn't able to learn it properly... now I need help... hehehe...

 

[pbandjay]

The easiest way that I know how to find this derivative is to re-write:
x = e^{\ln(x)}

So we have:
f(x) = x^{2x} = e^{2x\ln x}

f'(x) = \dfrac{d}{dx}e^{2x\ln x} = e^{2x\ln x}\dfrac{d}{dx}(2x\ln x) = x^{2x}(\frac{2x}{x} + 2\ln x) = 2x^{2x}(1 + \ln x)

 

[HallsofIvy]

pbandjay use the exponential form. What your teacher was talking about was this:
to differentiate x^2x, write y= x^2x and take the natural logarithm of both sides: ln(y)= ln(x^2x)= 2x ln(x). Now differentiate both sides. On the left, the derivative of ln(y), with respect to x, is (1/y) y'. On the right, use the product rule: (2x ln(x))'= (2) ln(x)+ (2x)(1/x)= 2ln(x)+ 2. So (1/y)y'= 2ln(x)+ 2 and y'= (2ln(x)+ 2)y= (2ln(x)+ 2)x^2x.


A rather amusing point: There are two obvious mistakes one could make in differentiating f(x)^g(x):

1) Treat g(x) as if it were a constant. Then you get (f(x)^g(x))'= g(x)(f(x))^g(x)-1f'(x)- which is, of course, wrong.

2) Treat f(x) as if it were a constant. Then you get (f(x)^g(x))'= f(x)^g(x) ln(f(x))g'(x)- which is also wrong.

To do it correctly, we can let y= f(x)^g(x) so that ln(y)= ln(f(x)^g(x))= g(x)ln(f(x) and differentiate both sides, using the product rule on the right.

(1/y) dy/dx= g'(x)ln(f(x)+ (1/f(x))g(x)f'(x) and, since y= f(x)^g(x),
dy/dx= (f(x)^g(x))'= f(x)^g(x)ln(f(x)g'(x)+ f(x)^g(x)-1f'(x),

the sum of the two mistaken derivatives!

 

[g_edgar]

the sum of the two mistaken derivatives!

part of the general method called "the chain rule for partial derivatives" which you (the OP) will meet later.

Another one:
Two mistaken ways to differentiate a product f(x) g(x) ... in one you think that f(x) is constant, in the other you think that g(x) is constant. BUT the product rule, the correct derivative, is just the sum of these two mistaken derivatives.

 

[pbandjay]

One problem I have always liked a lot that is very similar to this one is
y=x^{x^{x^{.^{.^{.}}}}}

To differentiate we can re-write: y = x^y and apply the logarithm.

y=x^y
Re-write: ln(y) = y ln(x)
Differentiate: y'/y = y/x + y' ln(x)
Re-arrange: y'/y - y' ln(x) = y/x ==> y'(1/y - lnx) = y/x ==> y'(1 - y lnx) = y²/x
Isolate y': y' = y²/[x(1 - y lnx)]

We have
y'= \frac{(x^{x^{x^{.^{.^{.}}}}})^2}{x(1-(x^{x^{x^{.^{.^{.}}}}})\ln x)}