I have a question about the following:
At 8:00 in the video below, the lecturer has shown that tan(θ) = the gradient, or tan(θ) = ∂y/∂x.
The lecturer then takes a derivative of both sides, 'in x'.
How is this possible, as the variable on the left is θ? I understand that the right hand side becomes the derivative of ∂y/∂x, which is ∂²y/∂x². I also get that the derivative of tan(x) = 1/cos²(x)
However, how does the derivative of tan(θ)=∂y/∂x, with respect to x become:
1/cos²(θ)*dθ/dx=∂²y/∂x²
Use the chain rule! \begin{align} \frac{\partial}{\partial x} \tan(\theta) &= \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta} \tan(\theta)\\ &= \frac{\partial \theta}{\partial x} \sec^2(\theta)\\ &= \frac{\partial \theta}{\partial x} \cdot \frac{1}{\cos^2(\theta)} \end{align} Notice that he writes $\frac{1}{\cos^2(\theta)}$, not $\frac{1}{\cos^2(x)}$.