Analyzing the Setup
Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the curtain on how calculus actually works. You have likely used the Chain Rule a thousand times to differentiate f(x)=sin(x2+1), but today, we are going to the source using the First Principle.
Imagine you are standing on a curve defined by f(x)=sin(x2+1). The derivative is the slope of the tangent line at any point x. We find this by taking a secant line connecting point P at x and a nearby point Q at x+h.
As h approaches zero, point Q slides toward P, and the secant line transforms into the tangent. This is the heartbeat of calculus.
The Algebraic Setup
We start with the formal definition of the derivative:
f′(x)=h→0limhf(x+h)−f(x)
Substituting our specific function, we obtain:
f′(x)=h→0limhsin((x+h)2+1)−sin(x2+1)
Do not panic at the complexity. First, expand the term (x+h)2 to get x2+2xh+h2. Our limit now takes the form:
f′(x)=h→0limhsin(x2+2xh+h2+1)−sin(x2+1)
The Trigonometric Transformation
To proceed, we must convert the difference of two sines into a product. We invoke the trigonometric identity:
sinC−sinD=2cos(2C+D)sin(2C−D)
Let C=x2+2xh+h2+1 and D=x2+1. When we calculate the difference C−D, the x2 and 1 terms vanish, leaving 2xh+h2. Dividing this by 2 yields h(x+2h).
When we calculate the sum C+D, we get 2x2+2xh+h2+2. Dividing by 2 yields x2+xh+2h2+1.
The Final Leap
Substitute these components back into our limit expression:
f′(x)=h→0limh2cos(x2+xh+2h2+1)sin(h(x+2h))
We now utilize the standard limit limθ→0θsinθ=1. Here, our angle is θ=h(x+2h).
To match the denominator, we multiply and divide by (x+2h). As h→0, the term h(x+2h)sin(h(x+2h)) approaches 1.
The cosine term simplifies to cos(x2+1), and the remaining factor (x+2h) simplifies to x. Multiplying the surviving pieces, we get 2⋅cos(x2+1)⋅x.
The final result is:
Look at that elegance. It matches the Chain Rule perfectly, proving that the shortcut you use every day is built on a foundation of pure, logical beauty.