Given that y = 3^x , find dy/dx in terms of x
Please help me! I know that you have to use ln but I am unsure as how to!
Thanks !
Core 4 question find dy/dx
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y=3^x
lny = ln3^x
lny=xln3
x= lny / ln3
dx= 1/ln3 * 1/y
dy
dx = 1/ yln3
dy
dy= ylna = (3^x) ln3
dx
whats with the neg? 
(Original post by Nerdatious)
Given that y = 3^x , find dy/dx in terms of x
Please help me! I know that you have to use ln but I am unsure as how to!
Thanks ! 
Can you explain the 4th step please?

(Original post by dean01234)
y=3^x
lny = ln3^x
lny=xln3
x= lny / ln3
dx = 1/ yln3
dy
dy= ylna = (3^x) ln3
dx 
(Original post by Nerdatious)
Can you explain the 4th step please?
I think it is just differentiation, but I haven't worked it out 
(Original post by dean01234)
Sorry, I'm not entireley sure what happens there... but that is the proof for (y=a^x =dy/dx = a^x lna) that I have copied down 
(Original post by Nerdatious)
Okay thank you, I will just learn that then Thanks for your help. 

(Original post by Nerdatious)
Sorry I don't understand, how would I go from there?
= d/dx (e^(x ln 3)) using the identity
= e^(x ln 3) ln 3 since ln 3 is a constant.
= 3^x ln 3 using the identity again 
(Original post by Nerdatious)
Can you explain the 4th step please?
Or just see the spoiler in my earlier post 
Find dy/dx for each of the following:
c.) y=xa^x  The CD answers doesn't show me how to do it at all and just tells me they have used the product rule
d.) y = 2^x / x  Then the quotient rule with this one. The CD again is just terrible and I need to know how to do these.
thanks 
(Original post by owen1994)
c.) y=xa^x  The CD answers doesn't show me how to do it at all and just tells me they have used the product rule
So
(Original post by owen1994)
d.) y = 2^x / x  Then the quotient rule with this one. The CD again is just terrible and I need to know how to do these.
Sp
Does that make any sense? I usually remember the two rules like this:
Product rule  "the first times the derivative of the second plus the second times the derivative of the first"
Quotient rule  "the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared"
Also, isn't this C3? That may be why they haven't explained it in the C4 book 
(Original post by Implication)
Product rule:
So
Product rule:
Sp
Does that make any sense? I usually remember the two rules like this:
Product rule  "the first times the derivative of the second plus the second times the derivative of the first"
Quotient rule  "the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared"
Also, isn't this C3? That may be why they haven't explained it in the C4 book 
If y = a^x
dy/dx = a^x ln a for any value of a.
Someone correct me if i'm wrong please, i'm saying this off the top of my head, but i'm pretty sure this is the equation we need to know.
Thanks for the negs, didnt know people hate maths that much. 
(Original post by Thasneemy)
If y = a^x
dy/dx = a^x ln a for any value of a.
Someone correct me if i'm wrong please, i'm saying this off the top of my head, but i'm pretty sure this is the equation we need to know. 
(Original post by owen1994)
Can you or any of the other people explain Question 4C in Edexcel Core 4, the newish book.
Find dy/dx for each of the following:
c.) y=xa^x  The CD answers doesn't show me how to do it at all and just tells me they have used the product rule
d.) y = 2^x / x  Then the quotient rule with this one. The CD again is just terrible and I need to know how to do these.
thanks
I dont know if you've solved these questions or whether you understand it now. But i'm going to try and help you through it. Forgive me if you understand the process, I can understand it may be annoying to be told the same thing over and over! Maybe by the time i've written this all out, someone's got there before me!
Ok, first general rule to keep in mind: if y = a^x then dy/dx= a^x lna
Question c: y=xa^x
Here, you have two things multiplying each other. you have x being multiplied by a^x. So immediately, you realise the need for product rule.
Product rule is: u(dv/dx) + v(du/dx)
First, lets establish what v is, what dv/dx is, what u is, what du/dx is.
v=x therefore dv/dx= 1
u=a^x therefore du/dx= a^xlna (using the first equation I showed you: y = a^x then dy/dx= a^x lna)
Ok, now, all you do is simply plug it into the product rule!
(x) x (a^xlna) + (1) x (a^x) > xa^xlna + a^x
Now, all you do is take out a common factor of a^x so you get:
a^x(xlna + 1) and that's done!
Question d: y = 2^x / x
Notice two things are being divided, so you immediately recognise the need for the quotient rule.
The quotient rule is: v(du/dx)  u(dv/dx) all divided by v^2.
Again, let's establish v, dv/dx, u and du/dx.
v= x therefore dv/dx= 1
u= 2^x therefore du/dx= 2^xln2 (using the same initial formula for powers)
Now, simply plug it all into the quotient rule!
((x) x (2^xln2)  (2^x)) / (x^2) (write this out on paper, it will look clearer.)
Then, you may take out from that a common factor of 2^x, leaving you with:
(2^x(xln2  1)) / x^2 and that's done!
Hope it makes sense! 
(Original post by electriic_ink)
Why not read the thread and find out? 
(Original post by owen1994)
I know all the rules from C3 and the rules for implicit equations I just can't get my head round these questions. I still don't really understand, but thanks for trying. I actually ended up just skipping them two questions and I moved on to finish the rest of differentiation. I only have Vectors and Integration to finish now, hopefully get vectors done this weekend . Thanks for help.
I don't know why you were negged for that either There seems to be a lot of negging going on in this thread aha (please don't get me too ) aha. Anyway, good luck.
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