How do I find the first derivative of a function using the power rule?

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To find the first derivative of the function f(x)=x^2-3x-3x^-2+5x^-3 using the power rule, the correct application involves differentiating each term separately. The initial attempt yielded f'(x)=2x-3+6x^-3-15x^-4, which was noted to have discrepancies in the exponents. The discussion clarified that the two forms of the derivative are equivalent despite looking different, and the proper approach to eliminate negative exponents is to multiply by x^4. This method simplifies the expression without needing actual division. Understanding these steps is crucial for mastering differentiation in calculus.
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f(x)=x^2-3x-3x^-2+5x^-3

I need help finding the 1st derivative of this function using the power rule.If you can help can you explain how you got the answer.I tried like five times but the differentiation calculator says I am getting the wrong answer.heres how o attempted it.

f(x)'=(2)(x)^2-1 - (3)(x)^1-1 - (3)(-2)(x)^-2-1 + (-3)(5)(x)^-3-1

here i am applying the power rule to each variable, coefficient and exponent group they each have a different color to help you identify them with the original function.

f(x)'=2x-3+6x^-3 -15x^-4
and this is the answer i get

this is the answer i should get

f(x)'=(2x^5-3x^4+6x-15)/(x^4)
 
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raeshun said:
f(x)=x^2-3x-3x^-2+5x^-3

I need help finding the 1st derivative of this function using the power rule.If you can help can you explain how you got the answer.I tried like five times but the differentiation calculator says I am getting the wrong answer.heres how o attempted it.

f(x)'=(2)(x)^2-1 - (3)(-2)(x)^1-1 - (3)(x)^-2-1 + (-3)(5)(x)^-3-1

ooo colors!
Notice the edit (in boldface) I have made. Do you see where you went wrong?
 
Sourabh N said:
ooo colors!
Notice the edit (in boldface) I have made. Do you see where you went wrong?

i fixed it.the coefficients are right but the exponents don't match up.
 
They do match for me. Can you show what you're getting? (and how you obtain it)
 
The two answers are exactly the same. Just do the indicated division.
 
You'll find this more and more in calculus. Your answer and the answer key may look different but still be equivalent.
 
HallsofIvy said:
The two answers are exactly the same. Just do the indicated division.

e^(i Pi)+1=0 said:
You'll find this more and more in calculus. Your answer and the answer key may look different but still be equivalent.

I'm sorry i am new to differentiation but how to i do the division?
 
You don't actually have to do any division. Multiply the numerator by x4 and multiply the denominator by x4. In other words, you're multiplying your answer by 1, to get rid of the negative exponents in the numerator.
 
Mark44 said:
You don't actually have to do any division. Multiply the numerator by x4 and multiply the denominator by x4. In other words, you're multiplying your answer by 1, to get rid of the negative exponents in the numerator.

It worked thanks :).but why do you multiply by x^4 instead of X^3 or something like that?
 
  • #10
raeshun said:
It worked thanks :).but why do you multiply by x^4 instead of X^3 or something like that?
... because you have an x-4 in you expression for the derivative.
 
  • #11
raeshun said:
It worked thanks :).but why do you multiply by x^4 instead of X^3 or something like that?

SammyS said:
... because you have an x-4 in you expression for the derivative.

Also because there is an x4 in the denominator in the answer. ;)
 
  • #12
raeshun said:
It worked thanks :).but why do you multiply by x^4 instead of X^3 or something like that?

SammyS said:
... because you have an x-4 in you expression for the derivative.

Sourabh N said:
Also because there is an x4 in the denominator in the answer. ;)
The only reason for multiplying by x4 over itself is because of the x-4 term in the numerator. The x4 is a result of this multiplication, not the reason for it.
 

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