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

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Answer 1 · 2018-01-12T11:09:46

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

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#