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asdfsystema
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deleted by accident .. ahh
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asdfsystema said:Please take a look and help me out. Simple derivatives that I'm confused with.
Thank you in advance :)
EDIT : Sorry Mark44. you're right. I put on two problems this time if that's okay with you . thank you !
Mark44 said:Speaking only for myself, I would be more inclined to help you on a single problem rather than a half dozen all at once.
and 2x/x^2= 2/x, of course: the derivative of 2log x is 2/x.try #2
log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5
or if it is this method ===== 2x/x^2 + 9log(x-3)-5log(x^2+3)
===== 2x/x^2+ 9*1/x-3 - 5* 2x/x^2+3
No, the derivative of 5ln(u) is NOT 5 ln(u)*du/dx, it is (5/u)*du/dx.Attempt at #2
f(x)= 5ln(5x+7ln(x)), find f'(x)
5ln(u)*du/dxu= 5x+7ln(x)) du/dx= 5+ 7*1/x ==== 5+7/x
5ln(5x+7ln(x))*5+7/x ?
f'(3)= 5ln(5(3)+7ln(3))*5+7/3 ??
Thank you very much ^^
asdfsystema said:thank you tim,
Attempt at #1
log[f(x)]= log[ (x^2)(x-3)^9 / (x^2+3)^5 ]
log[f(x)] = log(x^2)(x-3)^9 - log (x^2+3)^5 ?
try#1
log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5
i'm not sure if it is this ===== 2log(x) + 9log(x-3) - 5log(x^2+3)
try #2
log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5
or if it is this method ===== 2x/x^2 + 9log(x-3)-5log(x^2+3)
===== 2x/x^2+ 9*1/x-3 - 5* 2x/x^2+3
Attempt at #2
f(x)= 5ln(5x+7ln(x)), find f'(x)
5ln(u)*du/dx
u= 5x+7ln(x)) du/dx= 5+ 7*1/x ==== 5+7/x
5ln(5x+7ln(x))*5+7/x ?
f'(3)= 5ln(5(3)+7ln(3))*5+7/3 ??
asdfsystema said:f'(x)= (2/x + 9/x-3 -10x/x^2+3) * (x^2)(x-3)^9/(x^2+5)^5
f'(3)= (2/3+9/0 - 30/12) * (9)(0)^9/(9+5)^5
Um ... is f'(3) = 9 ? [/B]
I used d/dx 5ln(u) = 5/u * du/dx
so continuing off from there, 5/5x+7ln(x) * 5+7/x is the answer?
f'(3)= (5/5(3))+7ln(3) * 5+7/3
==== 1/3 + 7.6902 * 7.3333
final answer f'(3) = 56.7287 ?
Thanks , I tried to make it as clear as possible this time haha :)
Derivatives are mathematical tools used to measure the rate of change of a function. In science, they are crucial in studying the behavior and trends of various physical phenomena, such as the speed of a moving object or the growth of a population.
To find the derivative of a function, you can use the rules of differentiation, such as the power rule, product rule, and chain rule. These rules involve taking the limit of the function as the change in the independent variable approaches zero.
Sure, let's take the function f(x) = x^2. Using the power rule, the derivative of this function would be f'(x) = 2x. This means that for any given value of x, the slope of the tangent line to the curve at that point would be 2x.
Derivatives have numerous real-world applications in fields such as physics, engineering, economics, and biology. They are used to analyze and predict changes in physical systems, optimize processes, and model relationships between variables. For example, derivatives are used in calculating the maximum profit for a business or predicting future stock prices.
Yes, derivatives can be graphically represented as the slope of a tangent line to a curve at a specific point. On a graph, the derivative is shown as the slope of a line that just touches the curve at a particular point. The steeper the slope, the higher the value of the derivative.