# Derivatives and Chain Rules

I've recently been studying some basic calculus, and happened to find somewhat of a paradox.

Let's consider the three following functions :
$f(x) = 2x+5$
$g(x) = 3x-1$
$h(x) = f(g(x)) = 6x+3$

The differentials for the first two functions are :
$f'(x) = 2$
$g'(x) = 3$

However once we consider the differential for $h(x)$, things start to break down ...

If we use the chain rule form, we get :
$h'(x) = g'(x)\times f'(g(x)) = 3 \times 2 \times (3x-1) = 18x-6$

However, if we use the simpler linear form :
$h'(x) = 6$

Can somebody explain in simple, intuitive terms why using the chain rule form in this context does not work ?
By extension, what are the limitations and conditions in order to use the chain rule in calculus ?

Thanks

The notation $f'(g(x))$ doesn't mean “the derivative of $f$ multiplied by $g(x)$”, but

the derivative of $f$ computed at $g(x)$.

Since $f'(x)=2$ for every $x$, $f'(g(x))=2$ as well. Thus $$h'(x)=g'(x)f'(g(x))=3\cdot2=6. A more complex example can perhaps make you see better what's happening. Let f(x)=x^2+x and g(x)=2x^2-1. Then$$ h(x)=f(g(x))=(g(x))^2+g(x)=4x^4-4x^2+1+2x^2-1=4x^4-2x^2 $$so that h'(x)=16x^3-4x by direct computation. With the chain rule, consider f'(x)=2x+1 and g'(x)=4x, so$$ h'(x)=g'(x)\cdot f'(g(x))=4x\bigl(2g(x)+1\bigr) =4x(4x^2-1)=16x^3-4x $$Note that$$h(x) = f(g(x)) = 2(3x-1)+5=6x+3\implies h'(x)=6$$and$$h'(x)=f'(g(x))g'(x)=2 \cdot 3=6 $$indeed$$f(g(x))=2g(x)+5\implies f'(g(x))=\frac{df}{dg}=2

• @AllanRipley Let indicate $y=g(x)$ then $f(g(x))=f(y)=2y+5$ and $f'(g(x))=f'(y)=2$ since we are taking the derivative with respect to y. – gimusi Apr 13 '18 at 8:42

Because you made a mistake in calculation.

$f'(g(x))=2$, not $2\times(3x-1).$

There is almost no restriction to when you can apply chain rule. You only need that $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$.