Question

What is the derivative of $XY$ ?

Answer 1

To find the derivative we can use the slope formula, that is $slope=\dfrac{dy}{dx}$. $dy$ is the changes in $Y$and $dx$ is the changes in $X$.

To find the derivative of two variables in multiplication, this formula is used
$\dfrac{d}{dx}(uv)=u\dfrac{d}{dx}(v)+v\dfrac{d}{dx}(u)$
Mechanically, $\dfrac{d}{dx}(f(x))$ measures the rate of change of $f(x)$ with respect to $x$.Differentiation of any constant is zero. Differentiation of constant and a function is equal to constant times the differentiation of the function.

Complete step by step answer:
Let us derivate $XY$ with respect to $X$. Using the derivative formula,
$\dfrac{d}{dx}(uv)=u\dfrac{d}{dx}(v)+v\dfrac{d}{dx}(u)$
Substituting the terms
$\dfrac{d}{dX}(XY)=X\dfrac{d}{dX}Y+Y\dfrac{d}{dX}X$
The differentiation of $X$ with respect to $X$ is given by
$\dfrac{d}{dX}X=1$
As the derivation of the function is with respect to $X$ , the variable $Y$ cannot be differentiable. $Y$ is not constant.
The derivative of $XY$ with respect to $X$ is
$\dfrac{d}{dX}(XY)=X\dfrac{d}{dX}Y+Y$

Let us derivate $XY$ with respect to $Y$.Using the derivative formula,
$\dfrac{d}{dx}(uv)=u\dfrac{d}{dx}(v)+v\dfrac{d}{dx}(u)$
Substituting the terms
$\dfrac{d}{dY}(XY)=X\dfrac{d}{dY}Y+Y\dfrac{d}{dY}X$
The differentiation of $Y$ with respect to $Y$ is given by
$\dfrac{d}{dY}Y=1$
As the derivation of the function is with respect to $Y$, the variable $X$ cannot be differentiable. $X$ is not constant.
The derivative of $XY$ with respect to $Y$ is
$\therefore \dfrac{d}{dY}(XY)=X+Y\dfrac{d}{dY}X$

Note: $\dfrac{d}{dx}(f(x))=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$ is the formula for finding the derivative from the first principles. The slope is the rate of change of $y$ with respect to $x$ that means if $x$ is increased by an additional unit the change in $y$ is given by $\dfrac{dy}{dx}$ . Let us understand with an example, the rate of change of displacement of an object is defined as the velocity $Km/hr~$ that means when time is increased by one hour the displacement changes by $Km$. For solving derivative problems different techniques of differentiation must be known.