Hi Tom. How do I use this program to solve these two types of problems? What buttons do I push to locate the appropriate tools to use when solving these?
1.) F ‘ (x) for f(x) = x^2 – 6x + 4
2.) Find F ‘ (x) for f(x) = cubed root of x^2 + 1/x^2
I’m studying and practicing for a Calculus exam on Monday so I’m trying to learn and figure out your program today and tomorrow to assist me a little bit.
Thanks,
Nelson
F ‘ (x) for f(x) = x^2 – 6x + 4
I haven’t programmed this because it is too simple
I don’t think you need a program to solve this, and this is fundamental for your future calculus to learn, also easy
First you’ll notice that there are 3 items all seperated by minus and plus signs (x^2 -6x and +4)
for a derivative, you take the exponent 2 bring it in front of the ‘x’ variable and multiply it times anything in front (in this case 1), then subtract one from the exponent to get x^(2-1) = x^1 = x so 2x
Then put the minus sign down
and the derivative of -6x is just what’s infront of x or -6
the derivative of a number ( in this case 4) is zero so:
f ‘ = 2x-6
If the x^2 was 8x^2 then the exponent 2 would be multiplied by 8 to get 16 x
———————————————————–
Find F ‘ (x) for f(x) = cubed root of x^2 + 1/x^2
³√(x^2+1/x^2)
Anytime you see something like this you must get rid of the radical sign by the following:
(x^2+1/x^2)^(1/3) (the 3 in front of the radical is really ^1/3)
If the problem was like this:
²√(x^2+1/x^2) then it would be (x^2+1/x^2)^(1/2) (the 2 in front of the radical is really ^1/2)
Anytime you see an exponent (1/3) in calculus out side a function enclosed in parenthesis (x^2+1/x^2)
You think of the chain rule
You go to my menu ( index8() ) scroll down to chain rule and do the problem there
You have to press ALPHA first when you enter anything into my boxes for variables
Then:
(x^2+1/x^2)^(1/3)
Press ENTER
write everything you see down on your paper or test until the end answer
Pretty complicated problem and would not appear on a test, but maybe in homework, my programs are designed mainly for passing test questions.
Good luck, Tom
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