How do you differentiate #f(x) = 4/sqrt(tan^2(1-x) # using the chain rule?

1 Answer
Jan 5, 2016

Answer:

See the explanation section below.

Explanation:

To avoid the quotient rule, let's rewrite #f# as

#f(x) = 4(tan^2(1-x))^(-1/2)#

Now use the power and chain rules:

#f'(x) = 4[(-1/2)(tan^2(1-x))^(-3/2)] d/dx(tan^2(1-x))#

Finding #d/dx(tan^2(1-x))# will require the powe and chain rule applied to #(tan(1-x))^2#. So we get,

#f'(x) = 4[(-1/2)(tan^2(1-x))^(-3/2)][2(tan(1-x))]d/dx(tan(1-x))#

# = 4[(-1/2)(tan^2(1-x))^(-3/2)][2(tan(1-x))][sec^2(1-x)]d/dx(1-x)#

# = 4[(-1/2)(tan^2(1-x))^(-3/2)][2(tan(1-x))][sec^2(1-x)][-1]#

Simplifying algebraically, gets us,

#f'(x) = 4(tan(1-x))^(-3/2) tan(1-x)sec^2(1-x)#

# = (4tan(1-x)sec^2(1-x))/(tan^2(1-x))^(3/2)#

Avoiding Error

Remember that #(sqrtu)^2 = u#,

but if we square first, we get

#sqrt(u^2) = absu#

If you don't mind rewriting as a piecewise function, you could use:

#f(x) = 4/abs(tan(1-x)) = 4abscot(1-x) = {(4cot(1-x),1-x>0),(-4cot(1-x),1-x<0) :}#

Now differentiate each piece to get a piecewise derivative.