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# Differentiating inverse functions watch

1. I was looking over this example on differentiating inverse functions, and I really dont get how they get the answer from the second last line:
A function is defined as f (x) = 3e2x + 4
Find the derivative of the inverse function.

Solution
Let y = f-1 (x)
then f (y) = 3e^2y + 4 = x
and f ' (y) = 6e^2y

Again, using the rule for differentiating an inverse function, we have

d/dx (f^-1(x)) = 1/f'(y)

= 1/6e^2y

= 1/(2x -8)

Could anyone help (in plain English, lol)? I'm really having a bit of trouble getting to grips with this.
2. the differential of f-1(x) is 1/g'(x), where g(x) is f-1(x)

as d/dx (f^-1(x)) = 1/f'(y) and 1/f'(y) = 1/6e^2y :

f (f-1(x)) = x = f(y) = 3e2y + 4

so: ln|{x - 4}/3| = 2y
6e2y = 6eln|{x - 4}/3| = 6{x-4}/3 = 2x - 8 = f'(y)

so: 1/f'(y) = 1/(2x -8)
3. It might help to forget about the "f"s and just use x and y.

y = 3e^(2x) + 4

dy/dx = 6e^(2x)

dx/dy
= 1 / [dy/dx]
= 1 / (6e^(2x))
= 1 / (2y - 8)

Translating into function notation,

(d/dx)[f^(-1)(x)] = 1 / (2x - 8)
4. (Original post by Jonny W)
It might help to forget about the "f"s and just use x and y.

y = 3e^(2x) + 4

dy/dx = 6e^(2x)

dx/dy
= 1 / [dy/dx]
= 1 / (6e^(2x))
= 1 / (2y - 8)

Translating into function notation,

(d/dx)[f^(-1)(x)] = 1 / (2x - 8)
Thanks to both of you for replying so quickly. Jonny W, I'm probably being really thick here but i'm not sure about the lines in bold. how did you go from
1 / (6e^(2x)) to 1 / (2y - 8)?

I'm really bad at maths, sorry
5. (Original post by friendlyneutron)
Thanks to both of you for replying so quickly. Jonny W, I'm probably being really thick here but i'm not sure about the lines in bold. how did you go from
1 / (6e^(2x)) to 1 / (2y - 8)?

I'm really bad at maths, sorry
y = 3e^(2x) + 4
=> 2y = 6e^(2x) + 8
=> 6e^(2x) = 2y - 8
6. (Original post by Widowmaker)
y = 3e^(2x) + 4
=> 2y = 6e^(2x) + 8
=> 6e^(2x) = 2y - 8
I understand that step a bit better now thanks, but how can you go from 2y - 8 to 2x - 8. I think there's something fundamental i don't really get here.
7. Can no-one clarify it any further for me? I'd really appreciate it. I'll even resort to bribery! (i.e. rep )
8. Ohh. I like rep . He changes from y to x simply to make it look normal (it's standard to have x as the variable, not y). You've probably done it when finding inverses of functions. Say, find the inverse of y = e^x + 4. x = ln(y - 4) and so the inverse function is y = ln(x - 4) (switch x and y).
9. y = 3e^(2x) + 4

dy/dx = 6e^(2x)

dx/dy
= 1 / [dy/dx]
= 1 / (6e^(2x))

but y = 3e^(2x) + 4, so 2y = 6e^(2x) + 8, therefore (2y - 8) = 6e^(2x)
therefore (substituting in):
dx/dy = 1 / (2y - 8)

etc etc
10. (Original post by wacabac)
y = 3e^(2x) + 4

dy/dx = 6e^(2x)

dx/dy
= 1 / [dy/dx]
= 1 / (6e^(2x))

but y = 3e^(2x) + 4, so 2y = 6e^(2x) + 8, therefore (2y - 8) = 6e^(2x)
therefore (substituting in):
dx/dy = 1 / (2y - 8)

etc etc
How has that helped anyone?
You've just repeated what's been posted earlier.

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