How do you use the chain rule to differentiate #y=sinroot3(x)+root3(sinx)#?

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Answer 1 · 2018-03-29T11:13:03

#(dy)/(dx)=1/3[cosroot(3)x/root(3)(x^2)+cosx/root(3)(sin^2x)]#

Here,

#y=sinroot(3)x+root(3)sinx=sin(x^(1/3))+(sinx)^(1/3)#

#(dy)/(dx)=cosroot(3)xd/(dx)(x^(1/3))+1/3(sinx)^(1/3-1)d/(dx) (sinx)#

#=>(dy)/(dx)=cosroot(3)x(1/3x^(1/3-1))+1/3(sinx)^(-2/3)cosx#

#=1/3x^(-2/3)cosroot(3)x+1/3cosx(sinx)^(-2/3)#

#=1/3[cosroot(3)x/x^(2/3)+cosx/(sinx)^(2/3)]#

#=1/3[cosroot(3)x/root(3)(x^2)+cosx/root(3)(sin^2x)]#