I have the following problem: Let $M^n$ be an $n$-dimensional manifold and $f,g:M^n→R$ differentiable. Determine $d(fg)_p$ (the differential of $(fg)$ in the point $p$).
The result should be $d(fg)_p=d(f)g(p)+d(g)f(p)$.Is this correct? How can I show this? I tried to use a similar proof like the proof of the product rule. Can someone help me please?
Let $\gamma\colon]-\varepsilon,\varepsilon[\rightarrow M$ be a curve drawn on $M$ such that $\gamma(0)=p$ and let $\gamma'(0)=:v$, then by definition: $$\mathrm{d}f_p(v)=\frac{\mathrm{d}}{\mathrm{d}t}_{\vert t=0}f(\gamma(t)),\mathrm{d}g_p(v)=\frac{\mathrm{d}}{\mathrm{d}t}_{\vert t=0}g(\gamma(t)),\mathrm{d}(fg)_p(v)=\frac{\mathrm{d}}{\mathrm{d}t}_{\vert t=0}f(\gamma(t))g(\gamma(t)),$$ whence the result using the usual product rule applied to $f\circ\gamma,g\circ\gamma\colon]-\varepsilon,\varepsilon[\rightarrow\mathbb{R}$.