How do you differentiate #arc cot(-4tan(1/(1-3x^2)) )# using the chain rule?

1 Answer
Feb 4, 2018

Answer:

# d/dx "arccot"( -4tan(1/(1-3x^2)) ) = (24xsec^2(1/(1-3x^2)))/((1-3x^2)^(2) (1+16tan^2(1/(1-3x^2))) #

Explanation:

Let use define:

# y="arccot"( -4tan(1/(1-3x^2)) ) #

Then, we start with a table of standard derivatives:

# {: (ul("Function"), ul("Derivative"), ul("Notes")), (x^n, nx^(n-1), n " constant (Power rule)"), (tanx, sec^2x, ), ("arccot"^(-1)x, -1/(1+x^2), ), (f(g(x)), f'(g(x)) \ g'(x),"(Chain rule)" ) :} #

We can apply the chain rule to get:

# dy/dx = -1/(1+(-4tan(1/(1-3x^2)))^2) \ d/dx (-4tan(1/(1-3x^2))) #
# \ \ \ \ \ = 4/(1+16tan^2(1/(1-3x^2))) \ d/dx (tan(1/(1-3x^2))) #

Then, we apply the chain rule a second time:

# dy/dx = 4/(1+16tan^2(1/(1-3x^2))) \ sec^2(1/(1-3x^2)) \ d/dx(1/(1-3x^2)) #

# \ \ \ \ \ = (4sec^2(1/(1-3x^2)))/(1+16tan^2(1/(1-3x^2))) \ d/dx(1/(1-3x^2)) #

# \ \ \ \ \ = (4sec^2(1/(1-3x^2)))/(1+16tan^2(1/(1-3x^2))) \ d/dx((1-3x^2)^(-1)) #

Then, we apply the chain rule a third time:

# dy/dx = (4sec^2(1/(1-3x^2)))/(1+16tan^2(1/(1-3x^2))) \ (-1)(1-3x^2)^(-2) d/dx (1-3x^2) #

# \ \ \ \ \ = (4sec^2(1/(1-3x^2)))/(1+16tan^2(1/(1-3x^2))) \ (-1)/(1-3x^2)^(2) (-6x) #

# \ \ \ \ \ = (24xsec^2(1/(1-3x^2)))/((1-3x^2)^(2) (1+16tan^2(1/(1-3x^2))) #