Derivatives

[nobody]

Two problems are beating me up

1) f(x)= x/5 - 5/x


2) x^2/5 - 5/x^2

I don't know what rule to use first

Thanks in advance

 

[tkhunny]

Two problems are beating me up

1) f(x)= x/5 - 5/x


2) x^2/5 - 5/x^2

I don't know what rule to use first

Thanks in advance

You need only one, the Power Rule.

1) x/5 - 5*x^(-1)


2) x^(2/5) - 5*x^(-2)
or was that
2) (x^2)/5 - 5*x^(-2)

 

[nobody]

#2 is like the first you posted with x^(2/5) - 5/x^(-2) (2nd term is fraction)

On #1 what do i do with the fraction x/5

 

[nobody]

Is # 1
5+5x^-2

And # 2
2/5x^(-3/2) +10x^-3

 

[mark]

f(x)= x/5 - 5/x

f'(x) = d/dx (x/5) - d/dx(5/x) = 1/5 +5/x^2

use the quotient rule to find the derivitive of x/5

 

[nobody]

How did you get that answer?

 

[mark]

because of the difference rule

d/dx ( f(x) - g(x)) = d/dx(f(x)) - d/dx(g(x))

here f(x) = x/5 and g(x) = 5/x

by following the difference rule

f ' (x) = d/dx (x/5)- d/dx(5/x)

but x/5 is a quotent, so you use the quotent rule

which is (g(x) * f '(x) - f(x) * g'(x) ) / (g(x))^2

g(x) is the expression/number on the bottom
f(x) is the expression/number on the top

here f(x) = x
and g(x) = 5

so its 5 times the derivative of x - x times the direvitive of 5 all over 5 squared
and since the derivative of a constant is zero it becomes 5/5^2

then simply follow theses steps for 5/x

 

[nobody]

thanks so much for the help. Now I just need to figure out the second one.