Question

Find the derivative of the expression for an unspecified differentiable f(x)?

a) `f(x^2)`
b) `1/[1+f(x)]^2`
c) `(f(x)-1)/(f(x)+1)`

thanks in advance (:

Answer 1

Apply the appropriate rule in each part, using `f'(x)` when you need that derivative.

Explanation:

a) The derivative is `f'(x^2) * 2x` using the chain rule.

For part b) use either the quotient rule of rewrite it as `[1+f(x)]^-2` and use the power and chain rules.

For part c) use the quotient rule.

Answer 2

(a) `d/dx f(x^2) = 2x \ f'(x^2)`

(b) `d/dx 1/(1+f(x))^2 = (-2f'(x))/(1+f(x))^(3)`

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

Explanation:

Part (a):

By the chain rule, we have:

`d/dx f(x^2) = f'(x^2) d/dx x^2`
`" " = 2x \ f'(x^2)`

Part (b):

By the chain rule, we have:

`d/dx 1/(1+f(x))^2 = d/dx (1+f(x))^(-2)`
`" " = (-2)(1+f(x))^(-3) d/dx (1+f(x))`
`" " = (-2f'(x))/(1+f(x))^(3)`

Part (c):

By the quotient rule, we have:

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

`" " = ( (f(x)+1)f'(x) - (f(x)-1)f'(x) ) / (f(x)+1)^2`

`" " = ( f(x)f'(x)+f'(x) - f(x)f'(x)+f'(x) ) / (f(x)+1)^2`
`" " = ( 2f'(x) ) / (f(x)+1)^2`