Question

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

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
$\dfrac{df}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}.............(1.1)$
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
$\begin{aligned} & \dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{y(x+h)-y(x)}{h} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( u(x+h)+v(x+h) \right)-\left( u(x)+v(x) \right)}{h} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( u(x+h)-u(x) \right)-\left( v(x+h)+v(x) \right)}{h} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{u(x+h)-u(x)}{h}+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{v(x+h)-v(x)}{h}.............(1.3) \\\ \end{aligned}$
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
$\begin{aligned} & \dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{u(x+h)-u(x)}{h}+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{v(x+h)-v(x)}{h} \\\ & =\dfrac{du}{dx}+\dfrac{dv}{dx}.............(1.4) \\\ \end{aligned}$
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 $\underset{h\to 0}{\mathop{\lim }}\,(p+q)=\underset{h\to 0}{\mathop{\lim }}\,p+\underset{h\to 0}{\mathop{\lim }}\,q$ only if both $\underset{h\to 0}{\mathop{\lim }}\,p$ and $\underset{h\to 0}{\mathop{\lim }}\,q$ 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.