Shifting slope - a first order chain rule



Use a little geometry to explain why

\frac{d}{dx} f(x+c) = f'(x+c).

This is a special and simpler case of what we’ll later call the chain rule.

The derivative of the cosine it's geometric relationship with the sine

I like to imagine this problem in the form of a function where f(x) = x^2 which we all know how it would look on a graph. f(x+c) is just the function being shifted left or right depending on the value of c which is a constant. So c in my case could be 1, meaning f(x+1) is now equal to (x+1)^2. The “+1” shifts our x^2 function to the left as seen in this graph. (the green function is (x+1)^2 and red function is (x^2))

Now that we understand what the shift does to a normal function we can tackle what it does to a derivative of a function. We know that the derivative of x^2 is equal to 2x which is just a linear line. So we can use that to imagine what the derivative of (x+1)^2 is. By that same logic it should be a linear function, but only shifted over by that constant or as we know it c. In our example we can see the shift in the linear functions by a constant of 1. (the green function is d/dx(x+1)^2 and red function is d/dx(x^2))