Calculus

posted by .

differentiate each function

a) y = (cscx)(cotx)

b) y = x^2 + e^(2x)

• Calculus -

y = csc cot
y' = (-csc*cot) * cot - csc*csc^2
= -csc*cot^2 - csc^3
= -csc(csc^2+cot^2)

y = x^2+e^(2x)
y' = 2x + 2e^(2x)

Similar Questions

1. calculus

express in sinx 1 1 ---------- + -------- cscx + cotx cscx - cotx and express in cosx 1 + cot x ------- - sin^2x cscx = 1/sinx so what do i do w. that extra one on the top!?
2. trig

express this in sinx (1/ cscx + cotx )+ (1/cscx- cotx) i got 2sinx is that right?
3. drwls

My previous question: Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity. (secx/sinx)*(cotx/cscx) = (secx/cscx)(cotx/sinx) = (sinx/cosx)*cotx*(1/sinx) "The last steps should be obvious" Not to me. I can convert (sinx/cosx) to …
4. Pre-Calculus

Find a numerical value of one trigonometric function of x if tanx/cotx - secx/cosx = 2/cscx a) cscx=1 b) sinx=-1/2 c)cscx=-1 d)sinx=1/2
5. Precalculus

Simplify:[(cscx-cotx)(cscx+cotx)]/cscx
6. Math

prove:(cscx-cotx)^4 x (cscx+cotx)^4 = 1
7. Trigonometry

prove:(cscx-cotx)^4 x (cscx+cotx)^4 = 1
8. ap calc bc

hi! ok, i know that deriv(cscx) = -cscxcotx and that deriv(cos) = -sinx deriv(cotx) = -((cscx)^2) my question is: is this (statements below) correct?
9. trig

Prove (cscx+cotx)(cscx-cotx)=1