differentiate: (1)y= x^3/3+x^2/2+x/4. (2) y=2(1/x^2+x^2. (3)y=(2x^3 - 3/x^3)^1/3. (4) In (2x^3). (5) y=e^x/sinx

Question

differential calculus

Answer 1

y'=dy/dx:

(1) y'=x^2+x+1/4

(2) y'=2(x^-2+x^2)=2(-2x^-3+2x)=-4/x^3+4x=4(x-1/x^3)

(3) y'=(1/3)(2x^3-3/x^3)^(-2/3)(6x^2+9/x^4)=

(2x^3-3/x^3)^-(2/3)(2x^2+3/x^4)=

(2x^2+3/x^4)/(2x^3-3/x^3)^(2/3)=

(2x^2+3/x^4)/((2x^6-3)/x^3)^(2/3)=

(2x^2+3/x^4)/((2x^6-3)^(2/3)/x^2)=

(2x^4+3/x^2)/(2x^6-3)^(2/3)

(4) Let a=2^(1/3); y=ln(a^3x^3)=3ln(ax); y'=(3/ax)a=3/x

(5) y'=((e^x)sinx-(e^x)cosx)/(sinx)^2=(e^x)(cosecx-cotxcosecx)