// calculus  trig derivatives
posted by Laura .
how do i do i do y=cos(radt)+ rad(cost)?

f(x) = (3x^2) (csc4x)^4 // f'(x)=?
my notes say : f'(x) = (3x^2)5(csc4x)^4(csc(4x)(tan4x)(4)+(csc4x)^5(6)
I redid it and i got : f'(x) = (3x^2)5(csc4x)^4(csc4x)(cot4x)(4) + (csc4x)^5(6x)
which is correct? if its my notes why?
Respond to this Question
Similar Questions

Confused! PreCal
Verify that each equation is an identity.. tan A= sec a/csca I have notes (i wasn't here that day and teacher refuses to reteach) but I don't understand them here is the notes... Problem w/ same directions: Cos x= cotx/csc x = Cosx/Sin … 
Trigonometry
I need to prove that the following is true. Thanks. csc^2(A/2)=2secA/secA1 Right Side=(2/cosA)/(1/cosA  1) = (2/cosA)/[(1cosA)/cosA] =2/cosA x (cosA)/(1cosA) =2/(1cosA) now recall cos 2X = cos^2 X  sin^2 X and we could say cos … 
Calculus
yes! tnk u ok? It's actually (x>0.) Find the limit of cot(x)csc(x) as x approached 0? 
trig
how would you verify this trig identity (1+cos(x) / 1cos(x))  (1cos(x) / 1+cos(x)) = 4cot(x)csc(x) ? 
MathTrig
1. Create an algebraic expression for sin(arccosxarcsin3x) 2. The cosx=4/5, x lies in quadrant 4. Find sin x/2 3.Determine all solutions in (0,2pie) for sin4xsin^2x=3, cot^2v(31)cotx=v(3), cos^2x=cosx, and tan^2x6tanx+4=0 4. Solve; … 
Alg2/Trig
Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = 3/5. (Both u and v are in Quadrant II.) Find csc(uv). First of all, I drew the triangles of u and v. Also, I know the formula of sin(uv) is sin … 
calculus
f(x) = (3x^2) (csc4x)^4 // f'(x)=? my notes say : f'(x) = (3x^2)5(csc4x)^4(csc(4x)(tan4x)(4)+(csc4x)^5(6) I redid it and i got : f'(x) = (3x^2)5(csc4x)^4(csc4x)(cot4x)(4) + (csc4x)^5(6x) which is correct? 
Math
cos(tan + cot) = csc only simplify one side to equal csc so far I got this far: [((cos)(sin))/(cos)] + [((cos)(cos))/(sin)] = csc I don't know what to do next 
Calculus (Math)
Integral of (cot4x)^5 (csc4x)^7 
Trig verifying identities
I am having trouble with this problem. sec^2(pi/2x)1= cot ^2x I got : By cofunction identity sec(90 degrees  x) = csc x secx csc1 = cot^2x Then split sec x and csc1 into two fractions and multiplied both numerator and denominators …