Lack of rigour in usual derivation of Work-Energy Theorem The derivation of the Work-Energy theorem usually goes as follows:
You define the work done on a particle under net force $\vec{F}$ as
$$W=\int\limits_C \vec{F}\cdot\mathrm{d}\vec{r}$$
And then you use Newton's second law
$$\begin{align} W=\int\limits_C \vec{F}\cdot\mathrm{d}\vec{r}&=\int\limits_C m\vec{a}\cdot\mathrm{d}\vec{r}\\ 
&=\int\limits_C m\frac{\textrm{d}\vec{v}}{\textrm{d}t}\cdot\mathrm{d}\vec{r}=\int\limits_{t_i}^{t_f} m\frac{\textrm{d}\vec{v}}{\textrm{d}t}\cdot\frac{\mathrm{d}\vec{r}}{\textrm{d}t}\textrm{d}t=\int\limits_{t_i}^{t_f} m\frac{\textrm{d}\vec{r}}{\textrm{d}t}\cdot\frac{\mathrm{d}\vec{v}}{\textrm{d}t}\textrm{d}t \\
&=\int\limits_C m\frac{\textrm{d}\vec{r}}{\textrm{d}t}\cdot\mathrm{d}\vec{v}=\int\limits_C m\vec{v}\cdot\mathrm{d}\vec{v}
\end{align}$$
Up until here, I'm convinced. Non the less, the argument $$\textrm{d}(v^2)=\textrm{d}(\vec{v}\cdot\vec{v})=\vec{v}\cdot\textrm{d}\vec{v}+\vec{v}\cdot\textrm{d}\vec{v}=2\vec{v}\cdot\textrm{d}\vec{v}$$is often used. Then the integral resolves to $$W=\int\limits_{v_i}^{v_f}\frac{1}{2}m\textrm{d}(v^2)=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$My problem with that argument is that it is assumed that the exterior derivative works under the dot product just as it does under the normal product of real-valued functions. Can somebody explain to me (either geometrically or formally) why  $\textrm{d}(v^2)=\textrm{d}(\vec{v}\cdot\vec{v})=\vec{v}\cdot\textrm{d}\vec{v}+\vec{v}\cdot\textrm{d}\vec{v}=2\vec{v}\cdot\textrm{d}\vec{v}$ is true?
 A: One need not follow these steps. Indeed let $\gamma : I\subset \mathbb{R}\to \mathbb{R}^3$ be the trajectory of a particle. It's position at time $t$ is $\gamma(t)$, it's velocity is $\gamma'(t)$ and it's acceleration is $\gamma''(t)$. It's easy to see that
$$(\gamma'\cdot \gamma')'(t) = 2\gamma'(t)\cdot \gamma''(t),$$
so the work done by the resultant force because of Newton's Second Law can be written as
$$W = \int_\gamma \mathbf{F} = \int_I \mathbf{F}(\gamma(t))\cdot \gamma'(t)dt = \int_I m\gamma''(t)\cdot \gamma'(t)dt,$$
but as pointed out $\gamma''(t)\cdot\gamma'(t) = (\gamma'\cdot\gamma')'(t)/2$ so that
$$W = \dfrac{m}{2}\int_I(\gamma'\cdot\gamma')'(t)dt,$$
and by virtue of the fundamental theorem of calculus, if $I = [a,b]$ we have
$$W = \dfrac{m}{2}(|\gamma'(b)|^2-|\gamma'(a)|^2),$$
or setting $K(\mathbf{v}) = \dfrac{m}{2}|\mathbf{v}|^2$
$$W = K(\gamma'(b)) - K(\gamma'(a)) = \Delta K.$$
A: Formally, we have the standard dot product identity 
$$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{A}\cdot\mathbf{B}=\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\cdot \mathbf{B}+\mathbf{A}\cdot \frac{\mathrm{d}\mathbf{B}}{\mathrm{d}t}.$$
Inserting $\mathbf{A}=\mathbf{B}=\mathbf{v}$ gives 
$$\frac{\mathrm{d}\mathbf{v}^2}{\mathrm{d}t}=2\mathbf{v}\cdot \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}.$$
It then follows that 
$$\mathrm{d}\mathbf{v}^2\stackrel{\int}=2\mathbf{v}\cdot\mathrm{d}\mathbf{v}$$
by the chain rule.
A: Okay I have a simple to understand answer without requiring much knowledge. High school student can also understand this.
First of all the derivative of unit vector is perpendicular to the unit vector. This can be simply shown by considering a unit circle on a circle, then it changes by a very small amount. Now that small change will be along the end points of the vector. If they are very close, it will almost seem as if it is along the tangent (which is perpendicular to the radius) and also the unit vector. You can also think like this. Consider a particle performing non uniform circular motion. Clearly at any point, the velocity(derivative of position) will be along the tangent which is perpendicular to position vector.
Now x(vector) = magnitude of x vector* unit vector along x vector
Differentiate this equation wrt time to get
v(vector)= [speed * unit vector along x vector] + [magnitude of x vector* derivative of unit vector along x vector] this is because derivative of magnitude of x vector is speed
Now perform dot product operation on this equation with x vector.
LHS will become speed * magnitude of x vector only since the second term will cancel out due to dot product with perpendicular quantity
Multiplying dt on both sides we get
x(vector)•d(x(vector)) =xdx=dx²/2
This formula is valid for any vector quantity including v ie velocity
Sorry for writing like this because I don't know how to type vectors
