Question

y=x^x
dy/dx?

Answer 1

i know we are supposed to take logs on both sides but why?

Answer 2

@cwrw238

Answer 3

Use properties of logarithmic functions to expand the right side of the above equation as follows.

ln y = x ln x

We now differentiate both sides with respect to x, using chain rule on the left side and the product rule on the right.

y '(1 / y) = ln x + x(1 / x) = ln x + 1 , where y ' = dy/dx

Multiply both sides by y

y ' = (ln x + 1)y

Substitute y by x x to obtain

y ' = (ln x + 1)x x

Answer 4

what are you guys doing?

Answer 5

did anyone even understand the question?lol

Answer 6

d/dx (x^x) = (ln x +1)x^x

Answer 7

i need a HQ teacher here :p

Answer 8

whats wrong with my explanation?

Answer 9

I want to know "why" we took log on both the sides

Answer 10

you must to the same thing to both sides of the equation to keep it equal!

Answer 11

WHY DID WE USE LOGS BUT

Answer 12

u rnt getting my quest!

Answer 13

"The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 1000 is 10 to the power 3: 1000 = 10 × 10 × 10 = 10^3. More generally, if x = b^y, then y is the logarithm of x to base b, and is written y = log_b(x)"

Answer 14

oh gosh :|

Answer 15

leave it ! >.<

Answer 16

@hba @ghazi @Shadowys @sirm3d @hartnn someone :|

Answer 17

did you not cover logarithms in pre-calc? its difficult to ask someone to take a derivative of something they dont understand!

Answer 18

We MUST take ln to bring the exponent down, since we don't have the formula to take derivative of power in VARIABLE form :)

Answer 19

thats what I was looking for!
so ill use the same if i have powers like (2x,e^x,e^x+1 and stuff )?

Answer 20

Correct!

Answer 21

bingo!
Arigato!