Question

If u and v are differentiable functions of x and if $y=u+v$, then prove that
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{du}{dx}}+{\displaystyle \frac{dv}{dx}}$.

Answer 1

Hint: In this question, we are given that u and v are functions of x and they are differentiable. Therefore, we should use the definitions of derivative of a function and use this definition to find the derivative of y and express it in terms of the derivative of u and v as required in the question.

Complete step-by-step solution -
The derivative of a function f of x with respect to x is given by
${\displaystyle \frac{df}{dx}}=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}.............(1.1)$
As it is given that $$y=u+v$$ and y,u and v are functions of x, we should have
$$y(x)=u(x)+v(x)............(1.2)$$
For each value of x.
Thus, using y in place of f in equation (1.1), we obtain
$\begin{array}{rl}& {\displaystyle \frac{dy}{dx}}=\underset{h\to 0}{lim}{\displaystyle \frac{y(x+h)-y(x)}{h}}\\ & =\underset{h\to 0}{lim}{\displaystyle \frac{(u(x+h)+v(x+h))-(u(x)+v(x))}{h}}\\ & =\underset{h\to 0}{lim}{\displaystyle \frac{(u(x+h)-u(x))-(v(x+h)+v(x))}{h}}\\ & =\underset{h\to 0}{lim}{\displaystyle \frac{u(x+h)-u(x)}{h}}+\underset{h\to 0}{lim}{\displaystyle \frac{v(x+h)-v(x)}{h}}.............(1.3)\end{array}$
Now, we see that both the term in equation (1.3) are of the form of (1.1) but with f replaced by u and v respectively, therefore, we can rewrite equation (1.3) as
$\begin{array}{rl}& {\displaystyle \frac{dy}{dx}}=\underset{h\to 0}{lim}{\displaystyle \frac{u(x+h)-u(x)}{h}}+\underset{h\to 0}{lim}{\displaystyle \frac{v(x+h)-v(x)}{h}}\\ & ={\displaystyle \frac{du}{dx}}+{\displaystyle \frac{dv}{dx}}.............(1.4)\end{array}$
which is exactly as we wanted to prove in the question.

Note: We should note that in equation (1.3), we have assumed that u and v are differentiable functions because $\underset{h\to 0}{lim}(p+q)=\underset{h\to 0}{lim}p+\underset{h\to 0}{lim}q$ only if both $\underset{h\to 0}{lim}p$ and $\underset{h\to 0}{lim}q$ are well defined and exist. Thus, taking the limit of the sum as the sum of the limits in equation (1.3) will be valid only if both u and v are differentiable.