Calculus derivative problem?

Question

Hi. How do you differentiate this?

f(x)=((tan^2(x) - 1)(tan^4(x) + 10*tan^2(x) + 1)) / (3*tan^3(x))

The answer is f'(x) = 16*csc^4(2x) but I have no idea how to get it. This is for Calculus I. Thanks.

Answer 1

f(x) = (tan²x - 1)(tan⁴x + 10tan²x + 1) / (3tan³x)

Distribute terms using distributive property:  a( b + c) = ab + ac.

Simplify f(x) = (tan⁶x + 10tan⁴x + tan²x - tan⁴x - 10tan²x - 1) / (3tan³x)

f(x) = (tan⁶x + 9tan⁴x - 9tan²x - 1) / (3tan³x)

f(x) = tan⁶x / (3tan³x)+ 9tan⁴x / (3tan³x) - 9tan²x / (3tan³x) - 1/ (3tan³x)

f(x) = (1/3)tan³x + 3tanx - 3 / tanx - 1/ (3tan³x)

f(x) = (1/3)tan³x + 3tanx - 3 cotx - (1/3)cot³x

Apply derivative each side.

f'(x) = (1/3)(3tan²x)(sec²x) + 3(sec²x) - 3(-coses²x) - (1/3)(3cot²x)(-coses²x)

f'(x) = (tan²x)(sec²x) + 3(sec²x) + 3(coses²x) + ((cot²x)(coses²x)

Reciprocal identities: secx = 1/cosx and cosx = 1/sinx

Quotient Identities: tanx = sinx/cosx and cotx = cosx/sinx

f'(x) = (sin²x/cos²x)(1/cos²x) + 3(1/cos²x) + 3(1/sin²x) + (cos²x/sin²x)(1/sin²x)

f'(x) = (sin²x/cos⁴x) + 3/cos²x) + 3/sin²x) + (cos²x/sin⁴x)

Rewrite the expression with common denominotor.

f'(x) = (sin⁶x) + 3sin⁴xcos²x) + 3sin²xcos⁴x + (cos⁶x/) / (sin⁴xcos⁴x)

Principal algebraic expressions and formula: a⁶= (a²)³ = (a³)²  and (a + b)³ = a³ + 3a²b + 3ab² + b³

f'(x) = (sin²x)³ + 3(sin²x)²cos²x) + 3sin²x(cos²x )²+ (cos²x)³ / (sin⁴xcos⁴x)

f'(x) = (sin²x + cos²x )³ / (sin⁴xcos⁴x)

Pythagorean Identities: sin²x + cos²x = 1.

f'(x) = (1)³ / (sin⁴xcos⁴x)

Multiply nominator and denominator by 16.

f'(x) = (16) / (16)(sin⁴xcos⁴x)

f'(x) = (16) / (2sinxcosx)⁴

Double Angle Formula: 2sinAcosA = sin2A

f'(x) = 16 / (sin2x)⁴

f'(x) = 16[1/(sin2x)⁴]

Reciprocal identitie: 1/sinx = cscx

f'(x) = 16 csc⁴2x.