Logarithmic Derivative of 2^log2(x^2+1) with Base 2

[huzzi.123]

Homework Statement

2^log2(x^2+1)

2 after the log function is actually the base.

Homework Equations

derivative of log(x)= 1/x
derivative of a^x = a^x lna

The Attempt at a Solution

= 2^log2(x^2+1) ln2 *d[log2(x^2+1)]/dx
= 2^log2(x^2+1) ln2 [2x/(x^2+1)ln2]

ln2 in the numerator cancels with that in denominator but at the back of the book the answer written is x^2+1 which I'm not getting.

Regards

 

[Chestermiller]

Homework Statement

2^log₂(x^2+1)

2 after the log function is actually the base.

Homework Equations

derivative of log(x)= 1/x
derivative of a^x = a^x lna

The Attempt at a Solution

= 2^log2(x^2+1) ln2 *d[log2(x^2+1)]/dx
= 2^log2(x^2+1) ln2 [2x/(x^2+1)ln2]

ln2 in the numerator cancels with that in denominator but at the back of the book the answer written is x^2+1 which I'm not getting.

Regards

log₂(x^2+1) represents the power you have to raise 2 to to get (x^2+1).

In your problem, since 2 is being raised to this power in the expression 2^log₂(x^2+1), you must have that

2^log₂(x^2+1)=(x^2+1)

The derivative of this is just 2x. If you check your final result and cancel identical terms in the numerator and denominator, you are indeed left with 2x. So you solved the problem correctly (aside from reducing to lowest terms). If the book answer is (x^2+1), the book must be wrong.

 

[huzzi.123]

I'm not getting log2(x^2+1) in the denominator and thus can't cancel it.

I'm getting [2^log2(x^2+1) ln2 (2x)]/(x^2+1)ln2]

only ln2 can be cancelled.

 

[Chestermiller]

I'm not getting log2(x^2+1) in the denominator and thus can't cancel it.

I'm getting [2^log2(x^2+1) ln2 (2x)]/(x^2+1)ln2]

only ln2 can be cancelled.

We already said that 2^log2(x^2+1)=(x^2+1), so it cancels with the (x^2+1) in the denominator.

 

[Mark44]

= 2^log2(x^2+1) ln2 [2x/(x^2+1)ln2]

ln2 in the numerator cancels with that in denominator but at the back of the book the answer written is x^2+1 which I'm not getting.

Your notation is hard to read since you are using log2 to denote log₂ in one place, and ln2 to denote ln(2) in another.

Also, what you wrote above might not be what you meant. I'm going to assume that this is what you meant:
$$ 2^{log_2(x^2 + 1)} \cdot ln(2) \cdot \frac{2x}{(x^2 + 1) \cdot ln(2)}$$

The quantity in brackets that you wrote actually means ##\frac{2x}{x^2 + 1} * ln(2)##. Again I'm assuming that you meant that last ln(2) bit to be in the denominator, since you talked about cancelling them.

As already mentioned, ##2^{log_2(x^2 + 1)} = x^2 + 1##. In what you ended up with, everything cancels except the 2x, which is what you want to have left.

 

[huzzi.123]

Sorry for the confusion but that's how I could write.

So ln(2) in the denominator cancels with that in numerator. Also, when you apply the a^x formula it gives you;
2^log2(x^2+1) ln(2).

I should have this log term in the denominator to get it canceled but for I'm not getting it :/

 

[Chestermiller]

Sorry for the confusion but that's how I could write.

So ln(2) in the denominator cancels with that in numerator. Also, when you apply the a^x formula it gives you;
2^log2(x^2+1) ln(2).

I should have this log term in the denominator to get it canceled but for I'm not getting it :/

How many times do we have to say that 2^log2(x^2+1)=(x^2+1)? Does it cancel then?

 

[Mark44]

Sorry for the confusion but that's how I could write.

So ln(2) in the denominator cancels with that in numerator. Also, when you apply the a^x formula it gives you;
2^log2(x^2+1) ln(2).

What we've said several time is - don't leave it this way.

##2^{log_2(x^2 + 1)} = x^2 + 1##
There's a factor of x² + 1 in the numerator and the same factor in the denominator.

I should have this log term in the denominator to get it canceled but for I'm not getting it :/