How do you differentiate logarithms with different bases and exponents?

[DorumonSg]

How do you differentiate :

2^n/2?

You can't you just use the power rule?

The correct answer is 2^n/2 (In^2) 1/2

lg^2 x n

Where lg is log base 2.

The correct answer is 2lgn 1/n^2 . 1/n

Why is this so? Isn't lg^2 = 1? And differentiate n and we get 1?

Sorry if I sound lacking in maths, its not the main scope of my course, I just need it to process some algorithms.

 

[Mark44]

How do you differentiate :

2^n/2?

You can't you just use the power rule?

No, because the power rule doesn't apply. The power rule applies to functions for which the base is the variable and the exponent is a constant. Some examples include x², x⁵, and x^-1.

The function you're working with is an exponential function, one in which the base is constant but the exponent contains the variable.

I can't tell what your function is exactly. Is it
$$2^{n/2}$$

or

$$\frac{2^n}{2}$$

The correct answer is 2^n/2 (In^2) 1/2

Minor point: there is no such thing as "In" -- it's ln, "ell" "en" short for logarithmus naturalis, Latin for natural logarithm.

lg^2 x n

I have no idea what this is supposed to mean.

Where lg is log base 2.

The correct answer is 2lgn 1/n^2 . 1/n

Why is this so? Isn't lg^2 = 1? And differentiate n and we get 1?

Sorry if I sound lacking in maths, its not the main scope of my course, I just need it to process some algorithms.

 

[DorumonSg]

Its $$2^{n/2}$$

And oops my bad, I meant how do I differentiate lg^2 x n where lg is log base 2.

Since lg is log base 2 and its to the power of 2 wouldn't it be be 1? and if we differentiate n its 1?

 

[Mark44]

Its $$2^{n/2}$$

And oops my bad, I meant how do I differentiate lg^2 x n where lg is log base 2.

What does lg^2 x n mean? That "x" is throwing me off. If it means multiplication, then you are misunderstanding that logarithms are functions - log n does NOT mean that you multiply log times n.

Do you mean log₂(n)?

Here's the LaTeX for this function:
$$log_2(n)$$

Assuming that's what you meant, it's probably easiest to change to a different base.
$$log_2(n) = \frac{ln(n)}{ln(2)}$$

Differentiating both sides gives
$$d/dn(log_2(n)) = d/dn(\frac{ln(n)}{ln(2)}) = \frac{1}{n} \cdot \frac{1}{ln(2)}$$


Now, if you meant
$$(log_2(n))^2$$

you need to use the chain rule to differentiate this.

Since lg is log base 2 and its to the power of 2 wouldn't it be be 1? and if we differentiate n its 1?