Working the chain rule

Question

My daughter asked me for help with her college calculus and I am no help. Any assistance with showing her how to do this is greatly appreciated.

Find f(x) for f(x)= (5(x^2+2)cot(x)) / (3-cos(x)csc(x))

Answer 1

I would simplify it before taking the derivative. cos(x)csc(x) = cos(x)/sin(x) = cot(x)

f(x) = (5(x²+2)cot(x)) / (3-cot(x))

Multiply by tan(x)/tan(x)

(5(x²+2)cot(x)tan(x)) / ((3-cot(x))tan(x))

cot(x)tan(x) = 1

f(x) = (5(x²+2)/(3tan(x)-1)

Take the derivative using the quotient rule d(f(x)/g(x))/dx = (f'(x)g(x)-f(x)g'(x))/g²(x)

f '(x) =(5(2x)-5(x²+2)(3sec²(x))/(3tan(x)-1)² = (10x-15(x²+2)sec²(x))/(3tan(x)-1)²

Which could be simplified if needed.

Answer 2

if f(x)=p(x)/q(x)

f'(x)=[p'(x)q(x)-p(x)q'(x)]/(q(x))²

here p(x)=5(2+x²)cot(x)

p'(x)=5*[(2+x²)csc²(x)+(2x)cot(x)]

q(x)=(3-cot(x)),,,,,,cos(x)*csc(x)=cot(x)

q'(x)=-csc²(x)

edited =put in missing -1

f'(x)=(5[(2+x²)csc²(x)+(2x)cot(x)](3-cot(x))-[5(2+x²)cot(x)(-1)csc²x])/(3-cot(x))²