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

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.

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

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..

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.

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.