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# Derivative of logₐx (for any positive base a≠1)

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.C (LO)
,
FUN‑3.C.1 (EK)
Sal finds the derivative of logₐx (for any positive base a≠1) using the derivative of ln(x) and the logarithm change of base rule. He then differentiates log₇x and -3log_π(x).

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• When you take the derivative of log a (x), how do you know that 1/lna is a constant? Why isn't lnx a constant.
• Because a is a constant and the logarithm (of any base) of a constant number is a constant. Say a=3, 1/ln(3) is just a number while ln x is a function, as x is a variable
• At where is the proof for this log concept? in which lesson was it covered?
• At why don't we use the product rule for g(x) = -3 * log pi (x) or why is'nt -3 = 0 when grabbing the derivative of a constant
• You also could use the product rule, but because the derivative of a constant is 0, the first term "u'(x)*v(x)" in your result would disappear and you would be left with "u(x)*v'(x)" which is what we are doing anyway based on our differentiation rules.
• why the constant gets out of the d/dx ? for exemple d/dx[15 * log(x)] => 15 * d/dx[log(x)]
• The derivative of a constant times a function is equal to the constant times the derivative of the function. This "constant multiple rule" follows from the similar rule for limits.
• Why can he move the 1/ln(x) out of the derivative operator?
• Are you talking about what Sal did at ? He is moving `1/ln(a)`, which is a constant.
• is, xln(a) = ln(a^x)
• What if instead of having a logarithm in the form `log b (x)`, you have to differentiate a logarithm in the form `log x (a)` (where x is the variable which is to be differentiated with respect to and a is a constant). I was told by my Calculus teacher that this isn't really a thing, but I'm interested in how it would be done anyways. Is there a way to do this?
• By the change of base formula for logarithms, we can write logᵪa as ln(a)/ln(x). Now this is just an application of chain rule, with ln(a)/x as the outer function. So the derivative is -ln(a)/((ln(x))²)·(1/x).

Alternatively, we can use implicit differentiation: given y=logᵪ(a), we write x^y=a.
The left-hand side is e^(ln(x^y)), or e^(y·ln(x)). Differentiating both sides now gives e^(y·ln(x))·[y'ln(x)+y/x]=0.
The exponential is never 0, so we can divide it out to get y'ln(x)+y/x=0
y'ln(x)=-y/x
y'=-y/(x·ln(x))
Replace y from the original equation and get -logᵪ(a)/(x·ln(x)). Rewriting the numerator with change of base formula gives the same result as from the first method.

I don't know why your teacher told you this isn't a thing; it's a perfectly well-defined function.
• at why can't we just use the quotient rule for ln(x)/ln(a)
• We could, but ln(a) is a constant, so its derivative is 0. We would get
[ln(a)·ln'(x)-ln(x)·0]/(ln(a))²
=ln(a)·ln'(x)/(ln(a))²
=ln'(x)/ln(a)
=1/(xln(a))
• even log(a)/log(b) equals to loga(b). During the exercise, I got a problem where 7^x, and log(7)7^x was an incorrect choice. Why does only Ln work and not Log_10, even though they both do the same thing?