Calc
posted by Tammy .
Find the derivative of
a) y=cos (x/2)
b) yx3x^2 +6 = y^4 + x
a) sin (x/2)(2+x)/2^2
sin(x/2)(x/2)
b) x dy/dx + y  6x = 4y^3 dy/dx + 3x^2
dy/dx (x 4 y^3) = 3x^2 y +6x
dy/dx = (3x^2 y +6x)/(x 4 y^3)
I used inplicit differation for part b
are these correct? Thanks for your help in advance!
your first answer makes no sense.
you are simply finding the derivative of
y = cos (1/2)x
which is
y'=1/2 sin (x/2)
your second equation:
If your last term of the equation was a typo and it was x^3, then you are correct.
I like the way you showed the first line of your implicit derivative, but according to that line the last term must have been x^3 instead of x.
Respond to this Question
Similar Questions

trig
Reduce the following to the sine or cosine of one angle: (i) sin145*cos75  cos145*sin75 (ii) cos35*cos15  sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b)  sin(a)sin)(b) (1)The quantity … 
tigonometry
expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by b and using that cos(b)= cos(b) sin(b)= sin(b) gives: sin(ab) = sin(a)cos(b)  cos(a)sin(b) … 
algebra
Can someone please help me do this problem? 
calc
find the area between the xaxis and the graph of the given function over the given interval: y = sqrt(9x^2) over [3,3] you need to do integration from 3 to 3. First you find the antiderivative when you find the antiderivative … 
Mathematics  Trigonometric Identities
Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y … 
TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1  (3/4)sin^2 2x work on one side only! Responses Trig please help!  Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 … 
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 
calculus
Find the points on the curve y= (cos x)/(2 + sin x) at which the tangent is horizontal. I am not sure, but would I find the derivative first: y'= [(2 + sin x)(sin x)  (cos x)(cos x)]/(2 + sin x)^2 But then I don't know what to do … 
Calculus 12th grade (double check my work please)
1.)Find dy/dx when y= Ln (sinh 2x) my answer >> 2coth 2x. 2.)Find dy/dx when sinh 3y=cos 2x A.2 sin 2x B.2 sin 2x / sinh 3y C.2/3tan (2x/3y) D.2sin2x / 3 cosh 3yz...>> my answer. 2).Find the derivative of y=cos(x^2) … 
Trig
Find sin(s+t) and (st) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(1/5)Sin(3/5) = 0.389418 Sin(st) =sin(s)cos(t)  cos(s)sin(t) =sin(3/5)cos(1/5)  cos(1/5)sin(3/5) …