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# C3 differentiating e and ln help watch

Question:
(b) The curve C has the Cartesian equation y=3^(x-1)
By writing lny in terms of x, find an expression for dy/dx in terms of x.

y = 3x-1
so lny = (x - 1)ln3
1/y dy/dx = ln3
So dy = yln3 dx
= (3x-1)ln3

I understand the first stage but I'm not sure how they've got to 1/y dy/dx = ln3 and where the x-1 has gone.
2. (Original post by PuffyPenguin)

Question:
(b) The curve C has the Cartesian equation y=3^(x-1)
By writing lny in terms of x, find an expression for dy/dx in terms of x.

y = 3x-1
so lny = (x - 1)ln3
1/y dy/dx = ln3
So dy = yln3 dx
= (3x-1)ln3

I understand the first stage but I'm not sure how they've got to 1/y dy/dx = ln3 and where the x-1 has gone.
Note that then differentiate this w.r.t . Remember that is just a constant number.
3. (Original post by RDKGames)
Note that then differentiate this w.r.t . Remember that is just a constant number.
Okay.. but why is it 1/y dy/dx?
That make sense though, so thanks
4. (Original post by PuffyPenguin)
Okay.. but why is it 1/y dy/dx?
That make sense though, so thanks
Do you know how implicit differentiation works?

You essentially differentiate the LHS here w.r.t y then multiply the result by dy/dx.
5. (Original post by RDKGames)
Do you know how implicit differentiation works?

You essentially differentiate the LHS here w.r.t y then multiply the result by dy/dx.
No, I don't think I've covered implicit differentiation - thanks very much for the explanation
6. (Original post by PuffyPenguin)
No, I don't think I've covered implicit differentiation - thanks very much for the explanation
You can avoid implicit diff. by differentiating LHS w.r.t and the RHS w.r.t when you get to the line , which is the expected method you're supposed to use if you haven't covered implicit diff, I suppose.

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