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 (:

2 Answers
Jim H
Feb 5, 2018

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.

Steve M
Feb 5, 2018

(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#

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