How would I differentiate y = 4 ( 1/3 )^xWatch
I know the law of differentiating (1/3)^x but don't understand what to do with the 4.
e ^(x ln(1/3) + ln(4) )
But the ln(4) is stopping me..
u(x) = (1/3)^x
y(x) = 4u
dy/dx = 4 * du/dx
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.
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.
y(x) = 4u(x)
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".