Derivatives of exponential and logarithmic functions?

Question

My calculus class just covered this topic yesterday, but I don't understand it in the least.....

For example, f(x)=2lnx^3

would you take the derivative of 2 and then the derivative of lnx^3?

if so, then 2 would become zero and the whole thing cancels out?

Any help would be greatly appreciated.

Answer 1

What you've got there is a function of a function, so you need to use the old chain rule on it. (Remember also, that if F(x) = kf(x), where k is a constant, then F'(x) = kf'(x).)
(I take it x^3 means "x cubed".)
So, given that the derivative of ln y is 1/y, this makes the derivateive of 2 ln x^3 = (2/x^3) *3 x^2, and you can tidy up a bit and get (6x^2)/x^3 which cancels down to 6/x.

Answer 2

the by-made from ( [consistent] * function ) is [consistent] * function derivated so which you go away the two love that's and in simple terms derivate the function this i understand for specific, now the rest is precise on condition that i remember properly my calculus instructions if u is a function then ln(u) derivated is (u derrivated) / u and x ^ n derivated is: n * x ^ (n-a million) so on your case: f(x)= 2 ln(x ^ 3) the by-product is (via ' i mean derivated) f ' (x) = 2 * [ (x ^ 3) ' / (x ^ 3) ] (x ^ 3 ) derivated is 3x ^ 2 so f ' (x) = 2 * [ (3x ^ 2) / (x^3)] = 6 * (a million / x)

Answer 3

the derivative of
( [constant] * function )
is
[constant] * function derivated

so you leave the 2 like it is and just derivate the function


this i know for sure,
now the rest is accurate only if i remember correctly my calculus classes

if u is a function then ln(u) derivated is

(u derrivated) / u


and x ^ n derivated is: n * x ^ (n-1)

so in your case:

f(x)= 2 ln(x ^ 3)

the derivative is (by ' i mean derivated)

f ' (x) = 2 * [ (x ^ 3) ' / (x ^ 3) ]

(x ^ 3 ) derivated is 3x ^ 2 so

f ' (x) = 2 * [ (3x ^ 2) / (x^3)] = 6 * (1 / x)

Answer 4

2 is a constant and this 2lnx^3 becomes 6lnx, then 6 multiply by derivative of lnx. And it becomes 6/x. This is your answer.

Answer 5

derive ln x^3 derivative of ln y is dy/dt /y

here d(ln x^3) = 3x^2/x^3 = 3/x and to find your derivative you just multiply by 2

So your derivative is df(x) / dx = 6/x

a more easier way is to see that ln x ^3 = 3 ln x since derivative of lnx = 1/x , you find easily the same result