Chain rule: (dy)/(dx)=(dy)/(du)*(du)/(dx)dydx=dydu⋅dudx
We do this twice to derive both (x^2+5x)^2(x2+5x)2 and 2(x^3-5x)^32(x3−5x)3
d/(dx)(x^2+5x)^2ddx(x2+5x)2: Let u=x^2+5xu=x2+5x, then (du)/(dx)=2x+5dudx=2x+5
(dy)/(du)=2(x^2+5x)dydu=2(x2+5x)
So (dy)/(dx)=2(2x+5)(x^2+5x)dydx=2(2x+5)(x2+5x)
d/(dx)2(x^3-5x)^3ddx2(x3−5x)3: Let u=x^3-5xu=x3−5x, then (du)/(dx)=3x^2-5dudx=3x2−5
(dy)/(du)=6(x^3-5x)^2dydu=6(x3−5x)2
So (dy)/(dx)=6(3x^2-5)(x^3-5x)^2dydx=6(3x2−5)(x3−5x)2
Now adding both together,
(dy)/(dx)=2(2x+5)(x^2+5x)+6(3x^2-5)(x^3-5x)^2dydx=2(2x+5)(x2+5x)+6(3x2−5)(x3−5x)2