# math

posted by .

1)d/dx of cos4x/3-x^2(inculding quentient&chain rule)
2)sin^2x/1+cosx(same method)

• math -

#1

y = cos 4x/(3-x^2)

dy/dx = ( (3-x^2)(-4sin 4x) - cos 4x (-2x) )/(3-x^2)^2

try the second one, let me know what you got

• math -

you might be surprised at the answer to #2, unless you note that

sin^2/(1+cos) = (1-cos^2)/(1+cos) = 1-cos
no need for quotient rule or chain rule here.

## Similar Questions

1. ### Math, Calculus

Find the derivative of the function. y= xcosx - sinx What's the derivative?
2. ### calculus

Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the derivative
3. ### Math-trigonometry

Several questions from my homework, any are appreciated. Thanks! Solve: 1) sinx-tanx=0 2) 3csc^2x=4 Find all solutions within (0,2pi) 3) cos4x(cosx-1)=0 4) cos4x-7cos2x=8 Use inverse functions where necessary to find solutions from …
4. ### calculus

Let z = ∫(e^sin(t)dt) from x to y a = x b = y I tried thinking about it like a chain rule but even then i'm a little unsure. I know dz/dt = e^sin(t). Can you please point me in the right direction if i'm supposed to use the chain …
5. ### calculus

Let z = ∫e^(sin(t))dt from x to y a = x b = y I tried thinking about it like a chain rule but even then i'm a little unsure. I know dz/dt = e^(sin(t)). Can you please point me in the right direction if i'm supposed to use the …
6. ### calculus

Let z = ∫e^(sin(t))dt from x to y a = x b = y I tried thinking about it like a chain rule but even then i'm a little unsure. I know dz/dt = e^(sin(t)). Can you please point me in the right direction if i'm supposed to use the …
7. ### calculus

Find dz/dy and dz/dx Let z = ∫e^(sin(t))dt from x to y a = x b = y I tried thinking about it like a chain rule but even then i'm a little unsure. I know dz/dt = e^(sin(t)). Can you please point me in the right direction if i'm …
8. ### Precalculus/Trig

I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't know …
9. ### Calculus

Just got through with derivatives, power rule, and chain rule. I am supposed to find the first and second derivatives of the function. y = 4x/√x+1 and y = cos^2(x) y' = [(x+1)^(1/2)(4)]-[(4x)(1/2(x+1)^(-1/2))(1)] y' = [4(x+1)^(1/2)-4x]/[2(x+1)^(1/2)(x+1)] …
10. ### Mathematics-Integration

Question: For any positive integer n,show that integrate.[(sin x)^2n dx ] from o - π/2 = [(2n)!*π]/[(2)^(2n+1)*(n!)^2 ] What I thought: Let I =int.[(sinx)^2n dx] And again I= int.[ (sin(π/2-x))^2n dx] = int.[ (cos)^2n dx] 2I= int.[(sin …

More Similar Questions
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}