Find dy/dx for each of the following: Pls help?

Question

Find dy/dx for each of the following: Pls help?

a). x^2 -y^2 = 2x + y

b). y= sin^2xcos3x

c). y= 1+ sec x/tanx

d). y= x (ln cosec x)

Answer 1

a) 2x - 2yy' = 2 + y'

2yy' + y' = 2x - 2

y'(2y + 1) = 2(x - 1)

y' = 2(x - 1)/(2y + 1) ... (1)

If you want it to be in terms of x only, solve for y in x^2 -y^2 = 2x + y and substitute into (1).

b) Product Rule:

(sin^2 x)(-2sin3x) + (cos3x)(2sinx cosx)

c) Quotient Rule

d) Product Rule + Chain Rule

(x) (1/cscx * (-cscx cotx)) + (ln csc x) (1)

Answer 2

a)x^2-2x=y^2+y

x^2-2x+1/4=y^2+y+1/4

                =(y+1/2)^2

=>y+1/2=sqrt[x^2-2x+1/4]

differentiating we get

dy/dx= (1/2)[2x-2]/sqrt[x^2-2x+1/4]

b)y=sin^2xcos3x

dy/dx=2sinxcosxcos3x-3sinx^3x

         now expanding cos3x=3cosx-4cos^3x

    we get th final answer

c)is ity= 1+[secx/tanx]

=1+1/sinx

=1+cosecx

dy/dx=-cosecxcotx

   if  y= [1+secx]/tanx

        =[1+cosx]/sinx

        =cosecx+cotx

dy/dx=-cosecxcotx-cosec^2x

d)y=x(ln cosecx)

dy/dx=lncosecx+x(-cosecxcotx/cosecx)

       =ln cosecx-xcotx