I need to take the derivative of $f(x) = (x+c)⋅(x+c)$ with respect to time, where $x$ is a vector and $c$ is a constant vector. Is it correct to rewrite this as $f(x) = g(x)\cdot g(x)$ where $g(x)=x+c$ . And now using chain rule for dot product $f'(x) = 2g'(x)\cdot g(x) \frac{dx}{dt}$. Then we get $g'(x) = 1$. Resulting in $f'(x) = 2(x+c)x'$?
Take first derivative of dot product
derivatives
Best Answer
This is it. In fact, recall that for a vector $\mathbf{v}$, we have that $||\mathbf{v}||^2 = \mathbf{v} \cdot \mathbf{v}.$ Taking the derivative of this object is just using the chain rule.
Furthermore, the dot product obeys a product rule of differentiation similar to the scalar case: $(\mathbf{u}(t) \cdot \mathbf{v}(t))' = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}'.$