Chain Rule for Differentials

Question

I'm trying this problem from Lee's Smooth Manifolds, but I'm not sure where I'm making a mistake:

Problem: Let $M$ be a smooth manifold, and let $f,g\in C^{\infty}(M)$. If $J\subset\mathbb{R}$ containing the image of $f$, and $h:J\to\mathbb{R}$ is a smooth function, then $d(h\circ f)=(h'\circ f)df$.

My attempt: Let $X\in T_pM$ and $\frac{d}{dt}\in T_{f(p)}\mathbb{R}$. We want equality of $d(h\circ f)X$ and $(h'\circ f)dfX$

\begin{align} d(h\circ f)X&=d(f^*h)X\\&=f^*(dh)X\\&=dh(f_*X)\\&=dh((dfX)\frac{d}{dt})\\&=(dfX)dh(\frac{d}{dt})\\&=\frac{dh}{dt}dfX \end{align}

So there's an $f$ missing somewhere but running through that calculation, I don't see where I lost it.

Answer 1

In the beginning, it is better to track the base points of all the objects involved in order to make sure that everything "compiles" correctly. Here,

$$ d(h \circ f)|_p (X) = d(f^{*}(h))|_p (X) = (f^{*}(dh))|_p(X) = dh|_{f(p)}(df|_p(X)) = h'(f(p)) \cdot df|_p(X) = ((h' \circ f) \cdot df)|_p(X) $$

so you lost your $f$ by not keeping track how the pullback $f^{*}$ acts.