Please tell me how they got this derivatives

[neutron star]

Ok, I need help understanding this really bad. I don't get it. There were examples of derivatives in the book but I don't get how they got the answer. Here were the examples.

They all said to find the derivatives algebraically.

f(x)=x^3 at x=-2

The answer was 12.

f(x)=x^3+5 at x=1

The answer was 3.

How did they get these answers, I don't understand it, can someone help explain?

 

[rock.freak667]

ok for f(x)=xⁿ we know that f'(x)=nx^n-1

So in your first example f(x)=x³, so what is n and what is f'(x)? When you find f'(x) just put x=2 into your equation.

 

[neutron star]

Ah, that clears it up. A little confusing but I see that 3(-2)^3^-^1=12

Thanks.

But for the second one, are you not supposed to do anything with the +5?

 

[neutron star]

Okay, I have a weird one of this. It says find the derivative of g(x)=1/x at x=2 algebraically. So I got

g'(x)=1(2)^1^-^1 which is 2^0 which =1. Right? There were no exponents so I put 1, just want to make sure I am doing this right.

 

[rock.freak667]

Ah, that clears it up. A little confusing but I see that 3(-2)^3^-^1=12

Thanks.

But for the second one, are you not supposed to do anything with the +5?

for the second one, f(x)=x³+5, then f'(x)=d/dx(x³)+d/dx(5)

Okay, I have a weird one of this. It says find the derivative of g(x)=1/x at x=2 algebraically. So I got

g'(x)=1(2)^1^-^1 which is 2^0 which =1. Right? There were no exponents so I put 1, just want to make sure I am doing this right.

g(x)=1/x , so we can see this is not in the form xⁿ.But we know a rule in indices that says 1/a^m=a^-m

So now using this rule what is another way to write 1/x?

g(x) ?

 

[neutron star]

for the second one, f(x)=x³+5, then f'(x)=d/dx(x³)+d/dx(5)
g(x)=1/x , so we can see this is not in the form xⁿ.But we know a rule in indices that says 1/a^m=a^-m

So now using this rule what is another way to write 1/x?

g(x) ?

Ok, so it would be -1(2)^-^1^-^1, which is -2^-2 =1/4 right?

 

[rock.freak667]

Ok, so it would be -1(2)^-^1^-^1, which is -2^-2 =1/4 right?

remember -a^b means you take '-a' and raise it to the power of b

-(a)^b means you take 'a' and raise to the power of 'b' and then multiply by -1

 

[neutron star]

remember -a^b means you take '-a' and raise it to the power of b

-(a)^b means you take 'a' and raise to the power of 'b' and then multiply by -1

Ok so you mean 2^-1 which is 1/2 and then 1/2(-1) which is -1/2?

 

[rock.freak667]

Ok so you mean 2^-1 which is 1/2 and then 1/2(-1) which is -1/2?

no no, you have -(2)^-1-1=-(2)^-2=-(2^-2). So what number do you get now?

 

[neutron star]

-1/4.

 

[rock.freak667]

-1/4.

and that is correct.

 

[neutron star]

Cool, thanks. I see how that works.

 

[Elucidus]

remember -a^b means you take '-a' and raise it to the power of b

This is false. -a^b means -(a^b).

If you meant raising -a to the b^th power, you have to write (-a)^b.

--Elucidus