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Thu 28 Jul, 2005 04:45 am
How do I differentiate x^2e^-x(3x+5)^3?
I came up with ln (x^2e^-x)+ln (3x+5)^3 =
2 ln (xe^-x) + 3 ln (3x+5) =
2 (1/x e^-x) + 3 (5/3x+5) =
2/(xe^-x) + 15/(3x+5)
Am I off track? Should I keep going?
You need to learn some basics like the differentiation of products and the chain rule. Then you won't have to wonder separately about each new problem. I can't figure out why your teachers aren't teaching you this.
?
My teacher sucks.
Can anyone help?
Basic rules
The derivative of
f(x) g(x) = f(x) g'(x) + f'(x) g(x)
You can extend this also. So if you have
a(x) b(x) c(x), you can assign f(x) as a(x) and g(x) as [b(x) c(x)] This means you are going to have to apply the above rule a second time since you need the derivative of [b(x) c(x)] for part of the answer. It looks like this...
The derivative of a(x) b(x) c(x) is the derivative of a(x) [b(x) c(x)] =
a(x) [b(x) c(x)]' + a'(x) [b(x) c(x)] =
a(x) [ b(x) c'(x) + b'(x) c(x)] + a'(x) b(x) c(x)
You can use this to answer your original question since you want the derivative of the product of three functions of x.
engineer wrote:The derivative of
f(x) g(x) = f(x) g'(x) + f'(x) g(x)
You can extend this also. So if you have
a(x) b(x) c(x), you can assign f(x) as a(x) and g(x) as [b(x) c(x)] This means you are going to have to apply the above rule a second time since you need the derivative of [b(x) c(x)] for part of the answer. It looks like this...
The derivative of a(x) b(x) c(x) is the derivative of a(x) [b(x) c(x)] =
a(x) [b(x) c(x)]' + a'(x) [b(x) c(x)] =
a(x) [ b(x) c'(x) + b'(x) c(x)] + a'(x) b(x) c(x)
You can use this to answer your original question since you want the derivative of the product of three functions of x.
The question states specifically for me to use the chain rule of logarithmic differentation which is defined as:
d/dx [ln u(x)] = 1/u(x) du/dx
Your reply looks like basic chain and product rule. ??
foxy_yellow wrote:Your reply looks like basic chain and product rule. ??
He was showing you the product rule.
Logarithmic Differentation
I need to use logarithmic differentation, writing the problem to logs before getting the derivative.
Given a function f(x), you can take the logarithm of the function f, i.e.,
g(x) = ln f(x)
and differentiate the both sides of the above, i.e.,
g'(x) = f'(x)/f(x),
and hence you can find f'(x) as
f'(x) = g'(x) f(x).
It will surely work for your problem.
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