# How would I differentiate y = 4 ( 1/3 )^x

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#1
y = 4 ( 1/3 )^x ?

I know the law of differentiating (1/3)^x but don't understand what to do with the 4.

Thanks.
Last edited by seals2001; 4 weeks ago
0
4 weeks ago
#2
(Original post by seals2001)
y = 4 ( 1/3 )^x ?

I know the law of differentiating (1/3)^x but don't understand what to do with the 4.

Thanks.
It's simply a constant multiplier.
1
#3
I have tried to prove it by writing y in the form of e ^ ln(x) but I don't know what to do as I have got it to :

e ^(x ln(1/3) + ln(4) )

But the ln(4) is stopping me..
0
4 weeks ago
#4
(Original post by seals2001)
I have tried to prove it by writing y in the form of e ^ ln(x) but I don't know what to do as I have got it to :

e ^(x ln(1/3) + ln(4) )

But the ln(4) is stopping me..
It (the 4) just multiplies (1/3)^x in both the function and it's derivative.The

So if
u(x) = (1/3)^x
Then
y(x) = 4u
dy/dx = 4 * du/dx
Last edited by mqb2766; 4 weeks ago
0
#5
(Original post by mqb2766)
It (the 4) just multiplies (1/3)^x in both the function and it's derivative.The

So if
u(x) = (1/3)^x
Then
y(x) = 4u
dy/dx = 4 * du/dx
Is it not possible to answer by using e^ln(x) ?

Thanks.
0
4 weeks ago
#6
(Original post by seals2001)
I have tried to prove it by writing y in the form of e ^ ln(x) but I don't know what to do as I have got it to :

e ^(x ln(1/3) + ln(4) )

But the ln(4) is stopping me..
mqb's method is the obvious one here, but it's perfectly possible to finish it from where you're stuck:

If f(x) = x ln(1/3) + ln 4, then

(by the chain rule)

so if you can differentiate f you should have no problem.
0
4 weeks ago
#7
(Original post by seals2001)
Is it not possible to answer by using e^ln(x) ?

Thanks.
Yes, but it's more obscure. From first principles, differentiation is linear: double the function -> double the derivative.
0
#8
(Original post by mqb2766)
Yes, but it's more obscure. From first principles, differentiation is linear: double the function -> double the derivative.
It has taken me about 30 minutes to understand this intuitively. I finally get it through visualisation. If I have 2 points, and get the smallest possible distance between the 2 points, let's call it h, and now calculate the gradient, then I have a value of f for the derivative. If I multiply the graph by a multiple of 4, then the y coordinates all increase by a factor of 4, yet h is still the same, so there gradient is unchanged.
0
4 weeks ago
#9
(Original post by seals2001)
It has taken me about 30 minutes to understand this intuitively. I finally get it through visualisation. If I have 2 points, and get the smallest possible distance between the 2 points, let's call it h, and now calculate the gradient, then I have a value of f for the derivative. If I multiply the graph by a multiple of 4, then the y coordinates all increase by a factor of 4, yet h is still the same, so there gradient is unchanged.
It follows directly from the limit definition
y(x) = 4u(x)
Then
dy/dx = lim_(h->0) (4u(x+h)-4u(x))/h = 4 lim_(h->0) (u(x+h)-u(x))/h = 4 du/dx

The advantage is it works for any u(x), and obviously any (constant) multiplier.

As you say, graphically, multiplying a function by 4 makes the tangent 4 times "steeper".
Last edited by mqb2766; 4 weeks ago
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