I wanted to test my understanding of multivariable differentiation...so I made up a problem of my own. I wish to find the derivative(total derivative) of the following function:$f:\Bbb R^2 \to \Bbb R$ defined by

\begin{equation} f(x,y) = \int_{\sin(x)}^{\cos(y)}g(t)dt \end{equation}

where $g:\Bbb R \to \Bbb R$ is a continuous function. First I must show that $f$ is differentiable. Note that

\begin{equation} f(x,y) = \int_{\sin(x)}^{0}g(t)dt + \int_{0}^{\cos(y)}g(t) = \int_{0}^{-\sin(x)}g(t)dt + \int_{0}^{\cos(y)}g(t) \end{equation}

Now define $\psi:\Bbb R\to \Bbb R$ by $\psi(x) = \int_{0}^{x}g(t)dt$. Denote the projection of $\Bbb R^2$ onto the second coordinate by $\pi$. Then both $\psi$ and $\pi$ and differentiable. Now

\begin{equation} \int_{0}^{\cos(y)}g(t) = \psi(\cos(\pi(x,y))) \end{equation}

This shows that $\int_{0}^{\cos(y)}g(t)$ is differentiable. Similarly $\int_{0}^{-\sin(x)}g(t)dt$ is differentiable. Hence $f$ is differentiable. Now by chain rule and the fundamental theorem of calculus we have

\begin{equation} \left(\int_{0}^{\cos(y)}g(t)\right)' = \psi'(cos(\pi(x,y)))\times -sin(\pi(x,y))\times \begin{bmatrix} 1\\ 1 \end{bmatrix} = g(\cos(y))\times -sin(y)\times \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} -g(\cos(y))sin(y)\\ -g(\cos(y))sin(y) \end{bmatrix} \end{equation}

a similar computation will show that

\begin{equation} \left(\int_{0}^{\sin(x)} g(t)dt \right)' = \begin{bmatrix} -g(-\sin(x))cos(x)\\ -g(-\sin(x))cos(x) \end{bmatrix} \end{equation}

Hence the total derivative of $f$ will be(hopefully)

\begin{equation} \begin{bmatrix} -g(\cos(y))sin(y)-g(-\sin(x))cos(x)\\ -g(\cos(y))sin(y)-g(-\sin(x))cos(x) \end{bmatrix} \end{equation}

I kindly request anyone to check my computation and give suggestions for improvements

  • $\begingroup$ Apologies for the confusion, but by "total derivative" of a function $f:(x,y,t) \mapsto f(x(t),y(t))$ do you mean $\frac{\partial f}{\partial t} +\frac{\partial f}{\partial x}\frac{\mathrm{d} x}{\mathrm{d} t} +\frac{\partial f}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} t}$? $\endgroup$ – K.defaoite May 26 '20 at 18:40
  • $\begingroup$ I think you may not have gotten the size of the total derivative entirely right. Since it's a function from $\mathbb{R}^2 \to \mathbb{R}$, the matrix will be a $2 \times 1$ matrix. $\endgroup$ – Osama Ghani May 26 '20 at 18:48

I'm not sure what convention's you're using for the matrix; the way I learnt it is that if $f: \Bbb{R}^n \to \Bbb{R}^m$ then $f'(a)$ is an $m \times n$ matrix. In your case, a $1 \times 2$ matrix. But yes, your approach for proving differentiability of $f$ using $\psi$ and the projections along with the chain rule and Fundamental theorem of calculus is absolutely correct, and I believe it's the most efficient way of approaching the question.

By the way, I took only a brief glance at your minus signs etc, but they appear to be right, so good job.


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