Hi Cassie,
All of these problems can be solved by using the "power rule" for derivatives combined with the "sum and difference rule" and the "constant multiple rule":
The power rule says:
If f(x) = x^n then f'(x) = n*x^(n-1) [Here I'm using ^ to denote exponentiation--for example, 2^2 = 2 squared = 4]
For example, if f(x) = x^5, the derivative of f(x), or f'(x), is 5*x^4
Or, if f(x) = x^4, f'(x)=4*x^3
The constant multiple rule says:
If f(x) = a*g(x) where a is a constant and g(x) is another function,
then f'(x) = a*g'(x)
For example, if f(x) = 3*x^5, then 3 is a constant and g(x)=x^5.
So f'(x)=3*g'(x)=3*5*x^4=15*x^4
The sum and difference rule says:
If f(x) = h(x) + g(x), f'(x) = h'(x) + g'(x) or
if f(x) = h(x) - g(x), f'(x) = h'(x) - g'(x)
For example, if f(x) = x^5 + x^4, h(x) = x^5 and g(x)=x^4
So f'(x) = h'(x) + g'(x) = 5x^4 + 4x^3
Remember that 1/x^n = x^(-n) and √x = x^(1/2) and you can use just these rules to solve every problem you gave.
Let's do a quick example:
Suppose we have f(x) = 4x - 2/x + √x
We can split that up into three functions within the main function:
g(x)=4x
h(x)=2/x=2*x^(-1)
j(x)=√x=x^(1/2)
Now we're going to find the derivatives of each of those:
g'(x)=4 (constant multiple rule, derivative of x = 1)
h'(x)=2*(-1)*x^(-2) (constant multiple rule with a=2, power rule with n=-1)
j'(x)=(1/2)*x^(-1/2) (power rule with n=1/2)
To get the overall derivative, f'(x), now we're going to use the sum and difference rule:
f'(x) = g'(x) - h'(x) + j'(x) = 4 - 2*(-1)*x^(-2) + (1/2)*x^(-1/2)
Finally, we'll clean the expression up a bit to get:
f'(x) = 4 + 2/x^2 + 1/(2*√x)
Apply this same basic method to each of your problems and you'll have them done.
