How do you differentiate y=(x^2+1)root3(x^2+2)?

1 Answer
Jan 12, 2018

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

Explanation:

We have y=(x^2+1)(x^2+2)^(1/3)

y'=d/(dx)[(x^2+1)(x^2+2)^(1/3)]

color(white)(X)=d/(dx)[x^2+1] (x^2+2)^(1/3)+d/(dx)[(x^2+2)^(1/3)] (x^2+1)

color(white)(X)=(d/(dx)[x^2]+d/(dx)[1] )(x^2+2)^(1/3)+(d/(dx)[x^2+2] (x^2+2)^(1/3-1)*1/3) (x^2+1)

color(white)(X)=(2x+0)(x^2+2)^(1/3)+(((d/(dx)[x^2]+d/(dx)[2]) (x^2+2)^(-2/3))/3) (x^2+1)

color(white)(X)=2x(x^2+2)^(1/3)+(((2x+0) (x^2+2)^(-2/3))/3) (x^2+1)

color(white)(X)=2x(x^2+2)^(1/3)+((2x(x^2+2)^(-2/3))/3) (x^2+1)

color(white)(X)=2x(x^2+2)^(1/3)+(2x(x^2+1)(x^2+2)^(-2/3))/3

color(white)(X)=(6x(x^2+2)^(1/3))/3+(2x(x^2+1)(x^2+2)^(-2/3))/3

color(white)(X)=(6x(x^2+2)^(1/3)+2x(x^2+1)(x^2+2)^(-2/3))/3

color(white)(X)=(2x(3(x^2+2)^(1/3)+(x^2+1)(x^2+2)^(-2/3)))/3