# Calculus derivative problem?

+1 vote
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.
asked Jan 30, 2013 in CALCULUS

+1 vote

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.