Use the chain rule with Fundamental Theorem of calculus
g′(x)=f(h(x)) h′(x) − f(ϕ(x)) ϕ′(x)
=f(x²) × d/dx (x²) − f(5x+1) × d/dx (5x+1)
= (sin x²)/(x²) × (2x) - (sin (5x+1))/(5x+1) × 5
= (2x sin x²)/(x²) - (-5 sin (5x+1))/(5x+1)
How to find g′(x), when g(x)=∫ (sin t)/t dt from 5x+1 to x²?
