Solving derivative- possible without quotient rule?

[Persimmon]

Homework Statement

find the following derivative
d/dx [g(x + 1)(√(2+ (x + 8)^(1/3))/(cos(tan(sin(tan(sin x))))]
at x = 0

Homework Equations

The Attempt at a Solution

I split the big long derivative into 3 functions:
a(x) = g(x + 1)
b(x) = √(2+ (x + 8)^(1/3))
c(x) = cos(tan(sin(tan(sin x))))

and got : (a'x*bx + b'x*ax)(cx)-c'x*px*qx)/cx^2
subbing in x = 0 that simplified into 2*g'(1)
but it was mentioned that it's possible to solve this without the quotient rule. I can't figure out how you would solve this derivative if you didn't use the quotient rule.

 

[Mark44]

Homework Statement

find the following derivative
d/dx [g(x + 1)(√(2+ (x + 8)^(1/3))/(cos(tan(sin(tan(sin x))))]
at x = 0

Homework Equations

The Attempt at a Solution

I split the big long derivative into 3 functions:
a(x) = g(x + 1)
b(x) = √(2+ (x + 8)^(1/3))
c(x) = cos(tan(sin(tan(sin x))))

and got : (a'x*bx + b'x*ax)(cx)-c'x*px*qx)/cx^2
subbing in x = 0 that simplified into 2*g'(1)
but it was mentioned that it's possible to solve this without the quotient rule. I can't figure out how you would solve this derivative if you didn't use the quotient rule.

Instead of dividing by cos(...), you can multiply by sec(...). That way you're set up to use the product rule.

BTW, your notation is not very helpful.

a'x*bx + b'x*ax)(cx)-c'x*px*qx)/cx^2

As written this looks like a' * x * b * x etc. To be clearer, write a'(x) * b(x) etc.

 

[Persimmon]

Sorry for the unclear notation. I was trying to make it clearer so there wouldn't be endless brackets but I guess it's even more confusing that way. Thank you for your help!