Question #0fea1

1 Answer
Mar 28, 2017

(dy)/(dx)=((x^2-2)^2 (x (70 + x (173 + 37 x))-22))/(3 (4 + x)^(4/3) sqrt[1 + 2 x])

Explanation:

y=f_1(x)f_2(x)f_3(x) so

(dy)/(dx)=(df_1)/(dx)f_2 f_3+f_1((df_2)/(dx))f_3+f_1f_2((df_3)/(dx))

f_1=sqrt(2x+1) so (df_1)/(dx)=1/2 2/sqrt(2x+1)
f_2=(x^2-2)^3 so (df_2)/(dx)=3 xx 2 x(x^2-2)^2
f_3=1/root(3)(x+4)so (df_3)/(dx)=-1/3 1/root(4/3)(x+4)

Putting all together

(dy)/(dx)=(6 x sqrt[1 + 2 x] (x^2-2)^2)/(4 + x)^( 1/3) + (x^2-2)^3/((4 + x)^(1/3) sqrt[1 + 2 x]) - ( sqrt[1 + 2 x] (x^2-2)^3)/(3 (4 + x)^(4/3))

or

(dy)/(dx)=((x^2-2)^2 (x (70 + x (173 + 37 x))-22))/(3 (4 + x)^(4/3) sqrt[1 + 2 x])